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COPYRrCRT DEPOSnV 



MACHINE DESIGN 



BY 



ALBERT W. ^MITH 

Director of Sibley College, Cornell University 



AND 



GUIDO H. MARX 

Professor of Machine Design -f<^ 

Leland Stanford Junior University 



THIRD EDITION, REVISED AND ENLARGED 
FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS 

London: CHAPMAN & HALL, Limited 
1909. 



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Copyright, 1905, 1908, 1909, 

BY 

ALBERT W. SMITH and GUIDO H. MARX 



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PREFACE TO THE SECOND EDITION. 



One can never become a machine designer by studying books. 
Much help may come from books, but the true designer must 
have judgment, ripened by experience, in constructing and 
operating machines. One may know the laws that govern the 
development, transmission and application of energy; may have 
knowledge of constructive materials; may know how to obtain 
results by mathematical processes, and yet be unable to design 
a good machine. There is also needed a knowledge of many 
things connected with manufacture, transportation, erection and 
operation. With this knowledge it is possible to take results 
of computation and accept, reject and modify until a machine 
is produced that will do the required work satisfactorily. 

Professor John E. Sweet once said, ^' It is comparatively easy 
to design a good new machine, but it is very hard to design a 
machine that will be good when it is old." A machine must 
not only do its work at first, but must continue to do it with a 
minimum of repairs as long as the work needs to be done. The 
designer must be able to foresee the results of machine operation; 
he must have imagination. This is an inborn power, but it may 
be developed by use and by engineering experience. 

But there is a certain part of the designer's mental equipment 
that may be furnished in the class-room, or by books. This is 
the excuse for the following pages. Machine design cannot be 
treated exhaustively. There are too many kinds of machines 
for this and their differences are too great. In this book an 



IV PREFACE. 

effort is made simply to give principles that underlie all machine 
design and to suggest methods of reasoning which may be helpful 
in the designing of any machine. A knov^ledge of the usual 
university course in pure and applied mathematics is pre- 
supposed. 



INTRODUCTION, 



In general there are four considerations of prime importance 
in designing machines: I. Adaptation, II. Strength and Stiff- 
ness, III. Economy, IV. Appearance. 

I. This requires all complexity to be reduced to its lowest 
terms in order that the machine shall accomplish the desired 
result in the most direct way possible, and with greatest convenience 
to the operator. 

II. This requires the machine parts subjected to the action of 
forces to sustain these forces, not only without rupture, but also 
without such yielding as would interfere with the accurate action 
of the machine. In many cases the forces to be resisted may 
be calculated, and the laws of Mechanics and the known qualities 
of constructive materials become factors in determining propor- 
tions. In other cases the force, by the use of a ''breaking-piece," 
may be limited to a maximum value, which therefore dictates 
the design. But in many other cases the forces acting are neces- 
sarily unknown; and appeal must be made to the precedent of 
successful practice, or to the judgment of some experienced man, 
until one's own judgment becomes trustworthy by experience. 

In proportioning machine parts, the designer must always be 
sure that the stress which is the basis of the calculation or the 
estimate, is the maximum possible stress; otherwise the part 
will be incorrectly proportioned. For instance, if the arms of a 
pulley were to be designed solely on the assumption that they 



vi INTRODUCTION. 

endure only the transverse stress due to the belt tension, they 
would be found to be absurdly small, because the stresses resulting 
from the shrinkage of the casting in cooling are often far greater 
than those due to the belt pull. 

The design of many machines is a result of what may be called 
"machine evolution." The first machine was built according to 
the best judgment of its designer; but that judgment was falhble, 
and some part yielded under the stresses sustained ; it was replaced 
by a new part made stronger; it yielded again, and again was 
enlarged, or perhaps made of some more suitable material; it 
then sustained the applied stresses satisfactorily. Some other 
part yielded too much under stress, although it was entirely safe 
from actual rupture ; this part was then stiffened and the process 
continued till the whole machine became properly proportioned 
for the resisting of stress. Many valuable lessons have been learned 
from this process; many excellent machines have resulted from 
it. There are, however, two objections to it: it is slow and very 
expensive, and if any part had originally an excess of material, 
it is not changed; only the parts that yield are perfected. 

Modern analytical methods are rightly displacing it in all 
progressive establishments. 

III. The attainment of economy does not necessarily mean the 
saving of metal or labor, although it may mean that. To illustrate : 
Suppose that it is required to design an engine-lathe for the 
market. The competition is sharp; the profits are small. How 
shall the designer change the design of the lathes on the market 
to increase profits? {a) He may, if possible, reduce the weight 
of metal used, maintaining strength and stiffness by better dis- 
tribution. But this must not increase labor in the foundry or 
machine-shop, nor reduce weight which prevents undue vibrations. 
{h) He may design special tools to reduce labor without reduction 
of the standard of workmanship. The interest on the first cost 
of these special tools, however, must not exceed the possible gain 



INTRODUCTION. vii 

from increased profits. {c) He may make the lathe more con- 
venient for the workmen. True economy permits some increase 
in cost to gain this end. It is not meant that elaborate and 
expensive devices are to be used, such as often come from men 
of more inventiveness than judgment; but that if the parts can 
be rearranged, or in any way changed so that the lathes-man 
shall select this lathe to use because it is handier, when other 
lathes are available, then economy has been served, even though 
the cost has been somewhat increased, because the favorable 
opinion of intelligent workmen means increased sales. 

In {a) economy is served by a reduction of metal; in (6) by a 
reduction of labor; in {c) it may be served by an increase of both 
labor and material. 

The addition of material largely in excess of that necessary 
for strength and rigidity, to reduce vibrations, may also be in the 
interest of economy, because it may increase the durability of the 
machine and its foundation; may reduce the expense incident 
upon repairs and delays, thereby bettering the reputation of the 
machine and increasing sales. 

Suppose, to illustrate further, that a machine part is to be 
designed, and either of two forms, A or B, will serve equally well. 
The part is to be of cast iron. The pattern for A will cost twice 
as much as for B. In the foundry and machine-shop, however, 
A can be produced a very little cheaper than B- Clearly then if 
but one machine is to be built, B should be decided on; whereas, 
if the machine is to be manufactured in large numbers, A is 
prefera"ble. Expense for patterns is a first cost. Expense for 
work in the foundry and machine-shop is repeated with each 
machine. 

Economy of operation also needs attention. This depends 
upon the efficiency of the machine ; i.e., upon the proportion of the 
energy supplied to the machine which really does useful work. 
This efficiency is increased by the reduction of useless frictional 



viii INTRODUCTION. 

resistances, by careful attention to the design and means of lubri- 
cation of rubbing surfaces. 

In order that economy may be best attained, the machine 
designer needs to be familiar with all the processes used in the 
construction of machines — pattern -making, foundry work, forging, 
and the processes of the machine-shop — and must have them con- 
stantly in mind, so that while each part designed is made strong 
enough and stiff enough, and properly and conveniently arranged, 
and of such form as to be satisfactory in appearance, it also is 
so designed that the cost of construction is a minimum. 

IV. The fourth important consideration is Appearance. 
There is a beauty possible of attainment in the design of machines 
which is always the outgrowth of a purpose. Otherwise expressed, 
a machine to be beautiful must be purposeful. Ornament for 
ornament's sake is seldom admissible in machine design. And 
yet the striving for a pleasing effect is as much a part of the duty 
of a machine designer as it is a part of the duty of an architect. 

As a guiding principle, the general rule may be laid down 
that simplicity and directness are always best. Each member 
should be studied with strict reference to the function which it 
is to perform and the stresses to which it is subjected and then 
given the form and size best suited to meet the conditions with 
the greatest economy of material and workmanship. When 
combined, the parts must be modified in such manner as may be 
found necessary to the harmonious effect of the whole. 



CONTENTS. 



CHAPTER I. 

PAGE 

Preliminary i 

CHAPTER H. 
Motion in Mechanisms 14 

CHAPTER HI. 
Parallel or Straight-line Motions 39 

CHAPTER IV. 
Cams 47 

CHAPTER V. 
Energy in Machines 54 

CHAPTER VI. 
Proportions of Machine Parts as Dictated by Stress. 71 

CHAPTER VII. 
Riveted Joints 87 

CHAPTER VIII. 
Bolts and Screws 119 

CHAPTER IX. 

Means for Preventing Relative Rotation 148 

ix 



X CONTENTS. 

CHAPTER X. 

PAGE 

Sliding Surfaces 162 

CHiAPTER XI. 
Axles, Shafts, and Spindles , 170 

CHAPTER XII. 
Journals, Bearings, and Lubrication 181 

CHAPTER Xni. 
Roller- and Ball-bearings 210 

CHAPTER XIV. 
Couplings and Clutches 221 

CHAPTER XV. 
Belts 230 

CHAPTER XVI. 
Fly-wheels 256 

CHAPTER XVII. 
Toothed Wheels or Gears 275 

CHAPTER XVIII. 
Springs 349 

CHAPTER XIX. 
Machine Supports 354 

CHAPTER XX. 
Machine Frames • • 359 

Appendix c 395 

Index 403 



MACHINE DESIGN. 



CHAPTER I. 

PRELIMINARY. 

I. Definitions. — The study of machine design is based upon 
the science of mechanics, which treats questions involving the 
consideration of motion, force, work, and energy. Since it will 
be necessary to use these terms almost continually, it is well to 
make an exact statement of what is to be understood by them. 

Motion may be defined as change of position in space. 

A Force is one of a pair of equal, opposite, and simultaneous 
actions betw^een two bodies by which the state of their motion 
is altered, or a change in the form or condition of the bodies them- 
selves is effected. 

Work is the name given to the result of a force in motion. 

Energy is the capacity possessed by matter to do work. 

The law of Conservation of Energy underhes every machine 
problem. This law may be expressed as follows: The sum of 
energy in the universe is constant. Energy may be transferred 
in space; it may be stored for varying lengths of time; it may 
be changed from one of its several forms to another; but it can- 
not be created or destroyed. 

The application of this law to machines is as follows: A 
machine receives energy from a source, and uses it to do useful 
and useless work. 



2 MACHINE DESIGN. 

A complete cycle of action of a machine is such an interval 
that all conditions in the machine are the same at its beginning 
and end, each member of the machine having in the mean time 
gone through all motions possible to it. 

During a complete cycle of action of the machine, the energy 
received equals the total work done. The work done may appear 
as (a) useful work delivered by the machine, or as (b) heat due 
to energy transformed through frictional resistance, or as (c) 
stored mechanical energy in some moving part of the machine 
whose velocity is increased. The sign of the stored energy may 
be plus or minus, so that energy received in one cycle may be 
delivered during another cycle; but for any considerable time 
interval of machine action the algebraic sum of the stored energy 
must equal zero. 

For a single cycle: 
Energy received = useful work + useless work ± stored energy. 

For continuous action: 

Energy received = useful work + useless work. 
In operation a machine generally acts by a continuous repetition 
of its cycle. 

2. Efficiency of Machines. — In general, efficiency may be 
defined as the ratio of a result to the effort made to produce that 
result. In a machine the result corresponds to the useful work, 
while the effort corresponds to the energy received. Hence the 
efficiency of a machine = useful work ^energy received.* The 
designer must strive for high efficiency, i.e., for the greatest 
possible result for a given effort. 

3. Function of Machines. — Nature furnishes sources of 
energy, and the supplying of human needs requires work to be 
done. The function of machines is to cause matter possessing 
energy to do useful work. 

* The work and energy must, of course, be expressed in the same units. 



PRELmiNARY. 3 

The chief sources oj energy in nature available for machine 
purposes are: 

ist. The energy of air in motion {i.e.^ wind) due to its mass 
and velocity. 

2d. The energy of water due to its mass and motion or posi- 
tion. 

3d. The energy dormant in fuels which manifests itself as 
heat upon combustion. 

The general method by which the machine function is exer- 
cised may be shown by the following illustration: 

Illustration. — ^The water in a mill-pond possesses energy 
(potential) by virtue of its position. The earth exerts an attrac- 
tive force upon it. If there is no outlet, the earth's attractive 
force cannot cause motion; and hence, since motion is a neces- 
sary factor of work, no work is done. 

If the water overflows the dam, the earth's attraction causes 
that part of it which overflows to move to a lower level, and before 
it can be brought to rest again it does work against the force 
which brings it to rest. If this water simply falls upon rocks, 
its energy is transformed into heat, with no useful result. 

But if the water is led from the pond to a lower level, in a 
closed pipe which connects with a water-wheel, it will act upon 
the vanes of the wheel (because of the earth's attraction), and 
will cause the wheel and its shaft to rotate against resistance, 
whereby it may do useful work. The water-wheel is a machine 
and is called a Prime Mover, because it is the first link in the 
machine-chain between natural energy and useful work. 

Since it is usually necessary to do the required work at some 
distance from the necessary location of the water-wheel. Machinery 
of Transmission is used (shafts, pulleys, belts, cables, etc.), and 
the rotative energy is rendered available at the required place. 

But this rotative energy may not be suitable to do the re- 
quired work; the rotation may be too slow or too fast; a resist- 



4 MACHINE DESIGN. 

ance may need to be overcome in straight, parallel lines, or at 
periodical intervals. Hence Machinery of Application is intro- 
duced to transform the energy to meet the requirements of the 
work to be done. Thus the chain is complete, and the potential 
energy of the water does the required useful work. 

The chain of machines which has the steam-boiler and engine 
for its prime mover transforms the potential heat energy of 
fuel into useful work. This might be analyzed in a similar way. 

4. Free Motion. — The general science of mechanics treats of 
the action of forces upon ^'free bodies." 

In the case of a *' free body " acted on by a system of forces 
not in equilibrium, motion results in the direction of the resultant 
of the system. If another force is introduced whose line of 
action does not coincide with that of the resultant, the line of 
action of the resultant is changed, and the body moves in a new 
direction. The character of the motion, therefore, is dependent 
upon the forces which produce the motion. This is called free 
motion. 

Example. — In Fig. i, suppose the free body M to be acted 
on by the concurrent forces i, 2, and 3 
whose lines of action pass through the 
center of gravity of M. The line of 
action of the resultant of these forces is 
AB, and the body's center of gravity would 
move along this line. 

If another force, 4, is introduced, CD 
becomes the line of action of the resultant, 
and the motion of the body is along the line CD. 

5. Constrained Motion. — In a machine certain definite 
motions occur; any departure from these motions, or the pro- 
duction of any other motions, would result in derangement 
of the action of the machine. Thus, the spindle of an engine- 
lathe turns accurately about its axis; the cutting-tool moves 




PRELIMINAR Y. 5 

parallel to the spindle's axis; and an accurate cylindrical surface 
is thereby produced. If there were any departure from these 
motions, the lathe would fail to do its required work. In all 
machines certain definite motions must be produced, and all 
other motions must be prevented; or, in other words, motion 
in machines must be constrained. 

Constrained motion differs from free motion in being inde- 
pendent of the forces which produce it. If any force, not suffi- 
ciently great to produce deformation, be applied to a body whose 
motion is constrained, the result is either a certain predeter- 
mined motion, or no motion at all. 

6. Force Opposed by Passive Resistance. — A force may act 
without being able to produce motion (and hence without being 
able to do work), as in the case of the water in a mill-pond without 
overflow or outlet. This may be further illustrated: Suppose a 
force, say hand pressure, to be applied vertically to the top of a 
table. The material of the table offers a passive resistance, and 
the force is unable to produce motion, or to do work. 

It is therefore possible to offer passive resistance to such 
forces as may be required not to produce motion, thereby render- 
ing them incapable of doing work. Whenever a body opposes 
a passive resistance to the action of a force a change in its condi- 
tion is effected: the force sets up an equivalent stress in the 
material of the body. Thus, when the table offers a passive 
resistance to the hand -pressure, compressive stress is induced 
in the legs. In every case the material of the body must be of 
such shape and strength as to resist successfully the induced 
stress. 

In a machine there must be provision for resisting every 
possible force which tends to produce any but the required motion. 
This provision is usually made by means of the passive resistance 
of properly formed and sufficiently resistant metallic surfaces. 

Illustration I. — Fig. 2 represents a section and end view of 



MACHINE DESIGN. 



a wood-lathe headstock. It is required that the spindle, S, and 
the attached cone pulley, C, shall have no other motion than 




Fig. 2. 

rotation about the axis of the spindle. If any other motion is 
possible, this machine part cannot be used for the required pur- 
pose. At A and B the cylindrical surfaces of the spindle are 
enclosed by accurately fitted bearings or internal cylindrical sur- 
faces. Suppose any force, P, whose Hne of action lies in the 
plane of the paper, to be applied to the cone pulley. It may be 
resolved into a radial component, R, and a tangential component, 
T. The passive resistance of the cylindrical surfaces of the 
journal and its bearing, prevents R from producing motion; 
while it offers no resistance, friction being disregarded, to the 
action of T, which is allowed to produce the required motion, 
i.e., rotation about the spindle's axis. If the line of action of P 
pass through the axis, its tangential component becomes zero, 
and no motion results. If the line of action of P become tangen- 
tial, its radial component becomes zero, and P is wholly applied 
to produce rotation. If a force Q, whose line of action lies in 
the plane of the paper, be appHed to the cone, it may be resolved 
into a radial component, N, and a component, M, parallel to 
the spindle's axis. N is resisted as before by the journal and 
bearing surfaces, and M is resisted by the shoulder surfaces of 
the bearings, which fit against the shoulder surfaces of the cone 
pulley. The force Q can therefore produce no motion at all. 



PRELIMINARY. 



In general, any force applied to ' the cone pulley may be 
resolved into a radial, a tangential, and an axial component. 
Of these only the tangential component is able to produce motion; 
and that motion is the motion required. The constrainnient is 
therefore complete; i.e., there can be no motion except rotation 
about the spindle's axis. This result is due to the passive resist- 
ance of metallic surfaces. 

Illustration II. — R, Fig. 3, represents, with all details omitted, 
the "ram," or portion of a shaping-machine which carries the 



m m 


m 


R 


m.. 


^ , ^7 




; J 


•--r ; 


A 


H ' K 
B 



Fig. 3. 

cutting-tool. It is required to produce plane surfaces, and hence 
the "ram" must have accurate rectiHnear motion in the direction 
of HK. Any deviation from such motion would render the 
machine useless. 

Consider Fig. 3, ^. Any force which can be applied to the 
ram may be resolved into three components: one vertical, one 
horizontal and parallel to the paper, and one perpendicular to 
the paper. The vertical component, if acting upward, is resisted 
by the plane surfaces in contact at C and D ; if acting downward, 
it is resisted by the plane surfaces in contact at E. Therefore 
no vertical component can produce motion. The horizontal 
component parallel to the paper is resisted by the plane surfaces 
in contact at F or G, according as it acts toward the right or 
left. The component perpendicular to the paper is free to pro- 
duce motion in the direction of its line of action; but this is the 
motion required. 

Any force, therefore, which has a component perpendicular 
to the paper can produce the required motion, but no other 



8 MACHINE DESIGN. 

motion. The constrainment is therefore complete, and the 
result is due to the passive resistance offered by metalhc surfaces. 

Complete Constrainment is not always required in machines. 
It is only necessary to prevent such motions as interfere with 
the accompKshment of the desired result. 

The weight of a moving part is sometimes utilized to produce 
constrainment in one direction. Thus in a planer-table, and in 
some lathe-carriages, downward motion and unallowable side 
motion are resisted by metallic surfaces; while upward motion 
is resisted by the weight of the moving part. 

From the foregoing it follows that, as passive resistances 
can be opposed to all forces whose lines of action do not coincide 
with the desired direction of motion of any machine part, it may 
be said that the nature of the motion is independent of the forces 
producing it. 

Since the motions of machine parts are independent of the forces 
producing them, it follows that the relation of such motions may 
he determined without bringing force into the consideration. 

7. Kinds of Motion in Machines. — Motion in machines may 
be very complex, but it is chiefly plane motion. 

When a body moves hi such a way that any section of it re- 
mains in the same plane, its motion is called plane motion. All 
sections parallel to the above section must also remain, each in 
its o\vn plane. If the plane motion is such that all points of the 
moving body remain at a constant distance from some line, AB, 
the motion is called rotation about the axis AB. Example. — • 
A line -shaft with attached parts. 

If all points of a body move in straight parallel paths, the 
motion of the body is called rectilinear translation. Examples. — 
Engine cross-head, lathe-carriage, planer-table, shaper-ram. 
Rectilinear translation may be conveniently considered as a 
special case of rotation, in which the axis of rotation is at an 
infinite distance, at right angles to the motion. 

If a body moves parallel to an axis about which it rotates, 



PRELIMINARY, 



tne body is said to have helical or screw motion. Example. — • 
A nut turning upon a stationaiy screw. 

If all points of a body, whose motion is not plane motion, 
move so that their distances from a certain point, O, remain 
constant, the motion is called spheric motion. This is because 
each point moves in the surface of a sphere whose center is O. 
Example. — ^The arms of a fly-ball steam-engine governor, when 
the vertical position is changing. 

8. Relative Motion. — The motion of any machine part, like 
all known motion, is relative motion. It is studied by reference 
to some other part of the same machine. Some one part of a 
machine is usually (though not necessarily) fixed, i.e.^ it has no 
motion relative to the earth. This fixed part is called the frame 
of the machine. The motion of a machine part may be referred 
to the frame, or, as is often necessar}^, to some other part which 
also has motion relative to the frame. 

The kind and amount of relative motion of a machine part 
depend upon the motions of the part to which its motion is 
referred. 

Illustration. — Fig. 4 shows a press. A is the frame; C is a 
plate w^hich is so constrained that, its 
motion being referred to A, it may 
move vertically, but cannot rotate. 
Alotion of rotation is communicated to 
the screw B. The motion of B re- 
ferred to A is helical motion, i.e., 
combined rotation and translation. 
C, however, shares the translation of 
5, and hence there is left only rotation 
as the relative motion of B and C. FigT^! 

The motion of B referred to C is rotation. The motion of C re- 
ferred to B is rotation. The motion of C referred to A is translation. 

In general, if two machine members, M and AT", move relative 
to a third member, R, the relative motion of M referred to N 




TO MACHINE DESIGN. 

depends on how much of the motion of N is shared by M. If 
M and N hjive the same motions relative to R, they have no mo- 
tion relative to each other. 

Conversely, if two bodies have no relative motion, they have 
the same motion relative to a third body. Thus in Fig. 4, if 
the constrainment of C were such that it could share 5's rotation, 
as v^ell as its translation, then C would have helical motion rela- 
tive to the frame,^ and no motion at all relative to B. This is 
assumed to be self-evident. 

A rigid body is one in which the distance between elementary 
portions * is constant. No body is absolutely rigid, but usually 
in machine members the departure from rigidity is so slight that 
it may be neglected. 

Many machine members, as springs, etc., are useful because 
of their lack of rigidity. 

Points * in a rigid body can have no relative motion, and hence 
must all have the same motion. 

9. Instantaneous Motion and Instantaneous Centers or Cen- 
tros. — Points of a moving body trace more or less complex paths. 

If a point be considered as moving from 
one position in its path to another in- 
definitely near, its motion is called in- 
stantaneous motion. The point is mov- 
ing, for the instant, along a straight 
line joining the two indefinitely near 
together positions, and such a line is 
a tangent to the path. In problems which are solved by the 
aid of the conception of instantaneous motion it is only neces- 
sary to know the direction of motion; hence, for such purposes, 
the instantaneous motion oj a point is fully defined by a tangent to 
its path through the point. 

* In this volume these terms are used as interchangeable with the term 
■'particles" of mechanics. 




PRELIMINARY. II 

Thus in Fig. 5, if a point is moving in the path APB, when 
it occupies the position P the tangent TT represents its instan- 
taneous motion. Any number of curves could be drawn tangent 
to 2"r at P, and any one of them would be a possible path of 
the point; but whatever path it is following, its instantaneous 
motion is represented by TT. The instantaneous motion of a 
point is therefore independent of the form of its path. Any one 
of the possible paths may be considered as equivalent, for the 
instant, to a circle whose center is anywhere in the normal NN. 

In general, the instantaneous motion of a point, P, is equiva- 
lent to rotation about some point, O, in a line through the point P 
perpendicular to the direction of its instantaneous motion. 

Let the instantaneous motion of a point, A, Fig. 6, in a sec- 
tion of a moving body be given by the hne TT. Then the motion 
is equivalent to rotation about some point on the Hne ^5 as a 
center, but it may be any point, and hence the instantaneous 
motion of the body is not determined. But if the instantaneous 
motion of another point, C, be given by the line Pi Pi, this motion 
is equivalent to rotation about some point of CD. But the points 
A and C are points in a rigid body, and can have no relative 
motion, and must have the same motion, i.e., rotation about the 
same center. A rotates about some point of AB, and C rotates 
about some point of CD; but they must rotate about the same 
point, and the only point which 
is at the same time in both lines 
is their intersection, O. Hence 
A and C, and all other points 
of the body, rotate, for the instant, 
about an axis of which O is the 
projection; or, in other words, the 
instantaneous motion of the body 
is rotation about an axis of which 
O is the projection. This axis is the instantaneous axis of 




12 MACHINE DESIGN. 

the body's motion, and O is the instantaneous center of the 
motion of the section shown in Fig. 6. 

For the sake of brevity an instantaneous center will be called 
a centre. 

If TT and TiTi had been parallel to each other, AB and 
CD would also have been parallel, and would have intersected 
at infinity; in which case the body's instantaneous motion would 
have been rotation about an axis infinitely distant; i.e.y it would 
have been translation. 

The motion of the body in Fig. 6 is of course referred to a 
fixed body, which, in this case, may be represented by the paper. 
The instantaneous motion of the body relative to the paper is 
rotation about O. Let M represent the figure, and N the fixed 
body represented by the paper. Suppose the material of M 
to be extended so as to include O. Then a pin could be put 
through O, materially connecting M and N, without interfering 
with their instantaneous motion. Such connection at any other 
point would interfere with the instantaneous motion. 

The centra of the relative motion of two bodies is a pointy and 
the only one, at which they have no relative motion; it is a point, 
and the only one, that is common to the two bodies for the instant. 

It will be seen that the points of the figure in Fig. 6 might 
be moving in any paths, so long as those paths are tangent at 
the points to the lines representing the instantaneous motion. 

In general, centros of the relative motion of two bodies are 
continually changing their position. They may, however, remain 
stationary; i.e., they may become fixed centers of rotation. 

10. Loci of Centros, or Centrodes.* — As centros change posi- 
tion they describe curves of some kind, and these loci of centros 
may be called centrodes. 

* Centrode is here used in preference to "centroid," proposed by Professor 
Kennedy, because the latter term has grown to be generally accepted in mathe- 
matics as synonymous with "center of mass." 



PRELIMINARY. 13 

Suppose a section of any body, M, to have motion relatively 
to a section of another body, N (fixed), in the same or a parallel 
plane. Centros may be found for a series of positions, and a 
curve drawn through them on the plane of N would be the cen- 
trode of the motion of M relatively to N. If, now, M being 
fixed, N moves so that the relative motion is the same as before, 
the centrode of the motion of N relatively to M may be located 
upon the plane of M. Now, since the centro of the relative 
motion of two bodies is a point at which they have no relative 
motion, and since the points of the centrodes become succes- 
sively the centros of the relative motion, it follows that as the 
motion goes on, the centrodes would roll upon each other without 
slipping. Therefore, if the centrodes are drawn, and rolled 
upon each other without slipping, the bodies M and N will have 
the same relative motion as before. From this it follows that 
the relative plane motion of two bodies may be reproduced by 
rolling together, without slipping, the centrodes of that motion. 

II. Pairs of Motion Elements. — ^The external and internal 
surfaces by w^hich motion is constrained, as in Figs. 2 and 3, may 
be called pairs of motion elements. The pair in Fig. 2 is called 
a turning pair, and the pair in Fig. 3 is called a sliding pair. 

The helical surfaces by which a nut and screw engage with 
each other are called a twisting pair. These three pairs of 
motion elements have their surfaces in contact throughout. They 
are called lower pairs. Another class, called higher pairs, have 
contact only along elements of their surfaces. Examples. — Cams 
and toothed wheels. 



CHAPTER IL 

MOTION IN MECHANISMS. 

12. Linkages or Motion Chains; Mechanisms. 

In Fig. 7, b is joined to c by a turning pair; 
c '' d '' sliding " 

d '' a '' turning *' 

a '' b '' '' '' 




Fig. 7. 

Evidently there is complete constrainment of the relative 
majon of a, b, c, and d. For, d being fixed, if any motion occurs 
in either a, b, or c, the other two must have a predetermined 
corresponding motion. 

c may represent the cross-head, b the connecting-rod, and a 
the crank of a steam-engine of the ordinary type. If c were 
rigidly attached to a piston upon which the expansive force of 
steam acts toward the right, a must rotate about ad. This 
represents a machine. The members a, b, c, and d may be repre- 
sented for the study of relative motions by the diagram. Fig. 8. 

Thij assemblage of bodies, connected so that there is complete 

constrainment of motion, may be called a motion chain or linkage, 

and the connected bodies may be called links. The chain sho\vn 

is a simple chain, because no link is joined to more than two 

Qthers. If any links of a chain are joined to more than two 

14 



MOTION IN MECHANISMS. 



15 



Others, the chain is a compound chain. Examples will be given 
later. 

When one link of a chain is fixed, i.e.j when it becomes the 
standard to which the motion of the others is referred, the chain 
is called a mechanism. Fixing different links of a chain gives 
different mechanisms. Thus in Fig. 8, if d is fixed, the mechanism 
is that which is used in the usual type of steam-engine, as in 
Fig. 7. It is called the slider-crank mechanism. 

But if a is fixed, the result is an entirely different mechanism ; 
for h would then rotate about the permanent center ah, d would 
rotate about the permanent center ad, while c would have a more 
complex motion, rotating about a constantly changing centro, 
whose path may be found. 






\ 



\ - 



\ab I 




Fixing b or c would give, in each case, a still different mechan- 
ism. 

13. Location of Centros. — In Fig. 8 ^ is fixed and it is re- 
quired to find the centers of rotation, either permanent or in- 
stantaneous, of the other three links. The motion of a, relative 
to the fixed link d, is rotation about the fixed center ad. The 
motion of c relative to d is translation, or rotation about a centro 



1 6 MACHINE DESIGN. 

cd, at infinity vertically. The link b has a point in common 
with a; it is the centro, ab, of their relative motion. This point 
may be considered as a point in a or &; in either case it can have 
but one direction of motion relative to any one standard. As a 
point in a its motion, relative to d, is rotation about ad. For the 
instant, then, it is moving along a tangent to the circle through ab. 
But, as a point in b, its direction of instantaneous motion relative 
to d must be the same, and hence its motion must be rotation 
about some point in the line ad-ab, extended if necessary. Also, 
b has a point, be, in common with c; and by the same reasoning 
as above, be, as a point in b, rotates for the instant about some 
point of the vertical line through be. Now ab and be are points 
of a rigid body, and one rotates for the instant about some point 
of AB, and the other rotates for the instant about some point 
of CD] hence both ab and be (as well as all other points of b) 
must rotate about the intersection oi AB and CD. Hence bd 
is the centro of the motion of b relative to d. 

The motion of a may be referred to e (fixed), and ae will be 
found (by reasoning like that applied to b) to lie at the inter- 
section of the lines EF and GH. 

The motion chain in Fig. 8, as before stated, is called the 
slider-crank chain. 

14. Centres of the Relative Motion of Three Bodies are always 
in the Same Straight Line. — In Fig. 8 it will be seen that the 
three centros of any three links lie in the same straight line. 
Thus ad, ab, and bd are the centros of the links a, b, and d. This 
is true of any other set of three links. 

Proof. — Consider a, b, and d. The centro ab as a point in a 
has a direction of instantaneous motion relative to d perpen- 
dicular to a line joining it to ad. As a point in b it has a direc- 
tion of instantaneous motion relative to d perpendicular to a line 
joining it to bd. Therefore the lines ab-ad and ab-bd are both 
perpendicular to the direction of instantaneous motion of ab, and 



MOTION IN MECHANISMS. 17 

they also both pass through ah\ hence they must coincide, and 
therefore ah, ad, and hd must lie in the same straight line. But a, 
6, and d might be any three bodies whatever which have relative 
plane motion, and the above reasoning would hold. Hence it may 
be stated: The three centros of any three bodies having relative 
plane motion must lie in the same straight line. (The statement 
and proof of this important proposition is due to Prof. Kennedy.) 
15. Lever-crank Chain. Location of Centros. — Fig. 9 shows 
a chain of four links of unequal length joined to each other by 



jbd 




Fig. 9. 

turning pairs. The centros ah, ad, cd, and he may be located at 
once, since they are at the centers of turning pairs which join 
adjacent links to each other. The centros of the relative motion 
of h, c, and d are he, cd, and hd, and these must be in the same 
straight line. Hence hd is in the line B. The centros of the 
relative motion of a, h, and d are ah, hd, and ad; and these also 
must lie in a straight line. Hence hd is in the line A. Being 
at the same time in A and B, it must be at their intersection. 
By employing the same method ac may be found. 

16. The Constrainment of Motion in a linkage is independent 
of the size of the motion elements. As long as the cylindrical 
surfaces of turning pairs have their axes unchanged, the surfaces 
themselves may be of any size whatever, and the motion is un- 
changed. The same is true of sliding and twisting pairs. 



1 8 MACHINE DESIGN. 

In Fig. lo, suppose the turning pair connecting c and d to be 
enlarged so that it includes he. The link c now becomes a 




Fig. io. 

cylinder, turning in a ring attached to, and forming part of, the 
link d. he becomes a pin made fast in c and engaging with an eye 
at the end of h. The ccntros are the same as before the enlarge- 
ment of the pair cd, and hence the relative motion is the same. 

In Fig. II the circular portion immediately surrounding cd 
is attached to d. The link c now becomes a ring moving in a 
circular slot. This may be simplified as in Fig. 12, whence c 
becomes a curved block moving in a limited circular slot in d. 
The centros remain as before, the relative motion is the same, 
and the linkage is essentially unchanged. 

If, in the slider-crank mechanism, the turning pair whose 
axis is ah be enlarged till ad is included, as in Fig. 13, the motion 
of the mechanism is unchanged, but the link a is now called 
an eccentric instead of a crank. This mechanism is usually 
used to communicate motion from the main shaft of a steam-engine 
to the valve. It is used because it may be put on the main shaft 
anywhere without interfering with its continuity and strength. 

17. Slotted Cross-head. — The mechanism shown in Fig. 14 
is called the ''slotted cross-head meehanism.''^ Its centros may 
be found from principles already given. 



MOTION IN MECHANISMS. 



19 



This mechanism is often used as follows: One end of c, as 
£, is attached to a piston working in a cylinder attached to d. 
This piston is caused to reciprocate by the expansive force of 
steam or some other fluid. The other end of c is attached to 



Fig. II. 




\^> 



\ 






Fig. 12. 



another piston, which also works in a cylinder attached to d. 
This piston may pump water or compress gas (for example 
small ammonia compressors for refrigerating plants). The 
crank a is attached to a shaft, the projection of whose axis is 
ad. This shaft also carries a fly-wheel which insures approxi- 
mately uniform rotation. 



20 



MACHINE DESIGN. 



i8. Location of Centros in a Compound Mechanism. — It is 

required to find the centros of the compound linkage, Fig. 15. 
In any linkage, each link has a centro relatively to every other 




Fig. 13. 

link; hence, if the number of links = w, the number of centros = 
n(n — i). But the centro ab is the same as ba; i.e., each centro 



H 



__ uad 



/ 



3 



Fig. 14. 
is double. Hence the number of centros to be located for any 

linkage =—'^ . In the linkage Fig. 15, the number of centros 



6X5 



= 15.* 



* The links are a, b, c, d, e, and /. 
The centros: ab be cd de ef 
ac bd ce df 
ad be cf 
ae bf 



MOTION IN MECHANISMS. 



21 



The portion above the Hnk J is a shder-crank chain, and 
the character of its motion is in no way affected by the attachment 
of the part below d. On the other hand, the lower part is a 
lever-crank chain, and the character of its motion is not affected 
by its attachment to the upper part. The chain may therefore 
be treated in two parts, and the centros of each part may be 
located from what has preceded. Each part will have six centros, 
and twelve would thus be located, ad, however, is common to 









V'KllX I 



/7I 
/''I 

id/ ! 



/w 



i b\ ^J^ 









ae sr^j:^-— ' 



I 

I 




Fig. 15. 



the two parts, and hence only eleven are really found. Four 
centros, therefore, remain to be located. They are be, cj, bf, and 
ce. To locate be, consider the three links a, b, and e, and it 
follows that be is in the line A passing through ab and ae; con- 
sidering b, d, and e, it follows that be is in the line B through bd 
and de. Hence be is at the intersection of A and B. Similar 
methods locate the other centros. 

In general, for finding the centros of a compound linkage of 



22 MACHINE DESIGN. 

six links, consider the linkage to be made up of two simple chainS; 
and find their centros independently of each other. Then take 
the two links whose centro is required, together with one of 
the Hnks carrying three motion elements (as ^, Fig. 15). The 
centros of these links locate a straight line, A, which contains 
the required centro. Then take the two links whose centro is 
required, together with the other link which carries three motion 
elements. A straight Hne, B, is thereby located, which contains 
the required centro, and the latter is therefore at the intersection 
of A and B. 

19. Velocity is the rate of motion, or motion per unit time. 

Linear velocity is linear space moved through in unit time; 
it may be expressed in any units of length and time; as, miles 
per hour, feet per minute or per second, etc. 

Angular velocity is angular space moved through in unit time. 
In machines, angular velocity is usually expressed in revolutions 
per minute or per second. 

The linear space described by a point in a rotating body, or 
its Unear velocity, is directly proportional to _ its radius, or its 
distance from the axis of rotation. This is true because arcs 
are proportional to radii. 

If A and B are two points in a rotating body, and if n and r2 
are their radii, then the ratio of Hnear velocities 

linear veloc. A _ri 
linear veloc. B r? . 

This is true whether the rotation is about a center or a centro; 
i.e., it is true either for continuous or inrtantaneous rotation. 
Hence it applies to all cases of plane motion in machines ; because 
all plane motion in machines is equivalent to either continuous 
or instantaneous rotation about some point. 

To find the relation of linear velocity cf two poults in a machine 
member, therefore, it is only necessary to find \h^ relation of 



MOTION IN MECHANISMS. 23 

the radii of the points. The latter relation can easily be found 
when the center or centro is located. 

20. A vector quantity possesses magnitude and direction. It 
may be represented by a straight line, because the latter has 
magnitude (its length) and direction. Thus the length of a 
straight line, AB, may represent, upon some scale, the magnitude 
of some vector quantity, and it may represent the vector quantity's 
direction by being parallel to it, or by being perpendicular to it. 
For convenience the latter plan will here be used. The vector 
quantities to be represented are the linear velocities of points 
in mechanisms. The lines which represent vector quantities are 
called vectors. 

A line which represents the linear velocity of a point will 
be called the linear velocity vector of the point. The symbol of 
linear velocity will be VI. Thus VIA is the linear velocity of 
the point A. Also Va will be used as the symbol of angular 
velocity. 

If the linear velocity and radius of a point are known, the 
angular velocity, or the number of revolutions per unit time, 
may be found; since the Unear velocity -^ length of the circum- 
ference in which the point travels = angular velocity. 

All points of a rigid body have the same angular velocity. 

If the radii, and ratio of Unear velocities of two points, in 
different machine members are known, the ratio of the angular 
velocities of the members may be found as follows: 

Let ^4 be a point in a member M, and B a point in a member 

N. ri = radius of ^ ; ^2 = radius of B. VIA and VIB represent 

. VIA . 
the linear velocities of A and B, whose ratio, 777^, is known. 

VIA VIB 
Then VaA= and VaB= . 

27iri 271^2 

VoAVlA 27zr2_VlA ^2 VaM 
^^^^^ VaB ~ 2nn ^ VIB ~ VIB ^ n~ VaN ' 



24 MACHINE DESIGN. 

If M and N rotate uniformly about fixed centers, the ratio 

VaM 

77— r^ is constant. If either M or N rotates about a centro, the 
VaN ' 

ratio is a varying one. 

21. To find the relation of linear velocity of two points in 
the same link, it is only necessary to measure the radii of the 
points, and the ratio of these radii is the ratio of the Hnear veloci- 
ties of the points. 

In Fig. 16, let the smaller circle represent the path of Ay 

the center of the crank-pin of a slider-crank mechanism; the 

link d being fixed. Let the larger circle represent the rim of a 

pulley which is keyed to the same shaft as the crank. The 

pulley and the crank are then parts of the same link. The ratio 

VIA 
of velocity of the crank-pin center and the pulley surface =-t^ 

r 

=— . In this case the link rotates about a fixed center. The 

same relation holds, however, when the Hnk rotates about a 
centro. 




Fig. 16. 



22. Velocity Diagram of Slider-crank Chain. — In Fig. 17, 

, Vlah ab-bd ^ ... . , .i • 

the link d is fixed and yj^ = ^^_^^ . By similar triangles this 

expression is also equal to ~qZJ~' Hence, if the radius 0} the 



MOTION IN MECHANISMS. 



25 



crank circle he taken as the vector of the constant linear velocity 
of ah, the distance cut off on the vertical through O hy the line oj 
the connecting-rod {extended if necessary) will he the vector oj the 
linear velocity oj he. Project A horizontally upon hc-hd, locating 
B. Then hc-B is the vector of VI of the shder, and may be 

Fig. 17. 




Fig. 18. 



used as an ordinate of the linear velocity diagram of the shder. 
By repeating the above construction for a series of positions, 
the ordinates representing the VI of he for different positions of 
the shder may be found. A smooth curv^e through the extremi- 
ties of these ordinates is the velocity curve, from which the Vis 



26 MACHINE DESIGN. 

of all points of the slider's stroke may be read. The scale of 
velocities, or the linear velocity represented by one inch of ordi- 
nate, equals the constant linear velocity of ah divided by 0-ah 
in inches. 

23. Velocity Diagram of Lever-crank Chain. — It is required 
to find VI of he during a cycle of action of the mechanism shown 
in Fig. 18, d being fixed, and VI of ah being constant. The 
tv/o points ah and he may both be considered in the link h. 
All points in h move about hd relatively to the fixed link. 

Vlah ah-hd 

For most positions of the mechanism hd will be so located as to 
make it practically impossible to measure these radii, but a fine, 
as MN, drawn parallel to h cuts off on the radii portions which 
are proportional to the radii themselves, and hence proportional 
to the Vis of the points. Hence 

Vlah ah-M 
Vlhc ~ hc-N ' 

The arc in which he moves may be divided into any number of 
parts, and the corresponding positions of ah may be located. A 
circle through if, with ad as center, may be drawn, and the 
constant radial distance ab-M may represent the constant 
velocity of ah. Through Mi, M2, etc., draw Hnes parallel to the 
corresponding positions of h, and these lines will cut off on the 
corresponding line of c a distance which represents VI of he. 
Through the points thus determined the velocity diagram may 
be drawn, and the VI of he for a complete cycle is determined. 
The scale of velocities is found as in Sec. 22. 

24. The relation of Unear velocity of points not in the same 
link may also be found. 



MOTION IN MECHANISMS. 



27 



Required y,, , -^ referred to d as the fixed link, Fig. 19. 

The centre ah is a point in common to a and 6, the two links 
considered. Consider ah as a point in a; and its VI is to that 
of A as their radii or distances from ad. Draw a vector triangle 
with its sides parallel to the triangle formed by joining A, ah. 




Fig. 19. 

and ad. Then if the side Ai represent the VI of A^ the side fli&i 
will represent the VI of ah. Consider ah as a point in 6, and 
its VI is to that of B as their radii, or distances to hd. Upon 
the vector aihi draw a triangle whose sides are parallel to those 
of a triangle formed by joining ah, hd, and B. Then, from 
similar triangles, the side Bi is the vector of 5's linear velocity. 



Hence 



VI of A vector A 1 
VI oi 5"^ vector Bi 



The path of B during a complete cycle may be traced, and the 
VI for a series of points may be found, by the above method, then 
the vectors may be laid off on normals to the path through the 
points; the velocity curve may be drawn; and the velocity of 
B at all points becomes known. 



28 



MACHINE DESIGN. 



25. Angularity of Connecting-rod. — The diagram of VI of 
the slider-crank mechanism, Fig. 17, is unsymmetrical with 
respect to a vertical axis through its center. This is due to the 
angularity of the connecting-rod, and maybe explained as follows: 

In Fig. 20, ^O is one angular position of the crank, and BO 
is the corresponding angular position on the other side of the 
vertical through the center of rotation. The corresponding 
positions of the slider are as shown. But for position A the line 
of the connecting-rod, C, cuts off on the vertical through O a 
vector Oa, which represents the slider's velocity. For position 
B the vector of the slider's velocity is Ob and the velocity diagram 
is unsymmetrical. 




Fig. 20. 

If the connecting-rod were parallel to the direction of the 
slider's motion in all positions, as in the slotted cross -head 
mechanism (see Fig. 14), the vector cut off on the vertical through 
O would be the same for position A and position B and the 
velocity diagram would be symmetrical. 

Since the velocity diagram is symmetrical with a parallel 
connecting-rod and unsymmetrical with an angular connecting- 
rod, with all other conditions constant, it follows that the lack 
of symmetry is due to the angularity of the connecting-rod. 

The velocity diagram for the slotted cross-head mechanism 
is symmetrical with respect to both vertical and horizontal axes 
through its center. In fact, if the crank radius (= length of 



MOTION IN MECHANISMS. 29 

link a) be taken as the vector of the VI of ah, the linear velocity 
diagram of the slider becomes a circle whose radius =the length of 
the link a. Hence the crank circle itself serves for the linear velocity 
diagram, the horizontal diameter representing the path of the slider. 
26. Angularity of Connecting-rod, Continued. — During a por- 
tion of the cycle of the slider-crank mechanism, the slider's VI 
is greater that than of ab. This is also due to the angularity of 
the connecting-rod, and may be explained as follows: In Fig. 21, 
as the crank moves up from the position x, it will reach such a 
position, A, that the line of the connecting-rod extended will 
pass through B. OB in this position is the vector of the linear 
velocity of both ah and the slider, and hence their linear velocities 

«6B 




Fig. 21. 
are equal. When ah reaches B, the line of the connecting-rod 
passes through B\ and again the vectors — and hence the linear 
velocities — of ah and the slider are equal. For all positions 
between A and B the line of the connecting-rod will cut OB 
outside of the crank circle; and hence the linear velocity of the 
slider will be greater than that of ah. This result is due to the 
angularity of the connecting-rod, because if the latter remained 
always horizontal, its line could never cut OB outside the circle. 
It follows that in the slotted cross-head mechanism the maximum 
VI of the slider = the constant VI of ah. The angular space BOA, 
Fig. 21, throughout which VI of the slider is greater than the VI 
of ah, increases with increase of angularity of the connecting-rod; 
i.e., it increases with the ratio 

Length of crank 
Length of connecting-rod ' 



30 MACHINE DESIGN. 

27. Quick-return Mechanisms. — A slider in a mechanism 
often carries a cutting-tool, which cuts during its motion in one 
direction, and is idle during the return stroke. Sometimes the 
slider carries the piece to be cut, and the cutting occurs while 
it passes under a tool made fast to the fixed link, the return stroke 
being idle. 

The velocity of cutting is limited. If the limiting velocity 
be exceeded, the tool becomes so hot that its temper is drawn, 
and it becomes unfit for cutting. The limit of cutting velocity 
depends on the nature of the material to be cut. Thus annealed 
tool-steel and the scale surface of cast iron may be cut with 
carbon tool-steel at 10 to 20 feet per minute; wrought iron and 
soft steel at 25 to 30 feet per minute; while brass and the softer 
alloys may be cut at 40 or more feet per minute. With certain 
special tool-steels these speeds may be considerably exceeded. 
There is no limit of this kind, however, to the velocity during 
the idle stroke ; and it is desirable to make it as great as possible, 
in order to increase the product of the machine. This leads 
to the design and use of "quick-return" mechanisms. 

28. Slider-crank Quick Return, — If, in a slider-crank 
mechanism, the center of rotation of the crank be moved, so 
that the line of the slider's motion does not pass through it, the 
slider will have a quick-return motion. 

In Fig. 22, when the sHder is in its extreme position at the 
right, y4, the crank-pin center is at D. When the slider is at By 
the crank-pin center is at C. If rotation is as indicated by the 
arrow, then, while the slider moves from B to A, the crank-pin 
center moves from C over to D. And while the slider returns 
from A to B, the crank-pin center moves under from D to C. 
If the VI of the crank-pin center be assumed constant, the time 
occupied in moving from Z) to C is less than that from C to D. 
Hence the time occupied by the slider in moving from B io A 
is greater than that occupied in moving from A to B. The 



MOTION IN MECHANISMS, 3^ 

mean velocity during the forward stroke is therefore less than 
during the return stroke. Or the slider has a ''quick-return" 
motion. 




Fig. 22. 

It is required to design a mechanism of this kind for a length 
of stroke =-B^ and for a ratio 

mean VI forward stroke 5 
mean VI return stroke 7 * 

The mean velocity of either stroke is inversely proportional to 
the time occupied, and the time is proportional to the correspond- 
ing angle described by the crank. Hence 

mean velocity forward 5 angle /? 
mean velocity return 7 angle a 

It is therefore necessary to divide 360° into two parts which 
are to each other as 5 to 7. Hence a = 210" and ,/? = i5o°. Ob- 
viously l9 = i8o° — /? = 3o°. Place the 30° angle of a drawing 
triangle so that its sides pass through B and A. This condition 
may be fulfilled and yet the vertex of the triangle may occupy 
an indefinite number of positions. By trial O may be located so 
that the crank shall not interfere with the line of the slider.* 



*To avoid cramping of the mechanism, the angle BylZ> should equal or exceed 
i35°- 



32 MACHINE DESIGN. 

O being located tentatively, it is necessary to find the correspond- 
ing lengths of crank a and connecting-rod b. When the crank- 
pin center is at D, AO=b-a; when it is at C, BO=b + a. AO 
and BO are measurable values of length; hence a and b may 
be found, the crank circle may be drawn, and the velocity dia- 
grams may be constructed as in Fig. 17; remembering that the 
distance cut off upon a vertical through O, by the line of the 
connecting-rod, is the vector of the VI of the slider for the corre- 
sponding position when the VI of the crank-pin center is repre- 
sented by the crank radius. 

It is required to make the maximum velocity of the forward 
stroke of the slider = 20 feet per minute, and to find the corre- 
sponding number of revolutions per minute of the crank. The 
maximum linear velocity vector of the forward stroke = the 
maximum height of the upper part of the velocity diagram; 
call it Vh. Call the linear velocity vector of the crank-pin center 
F/2 = crank radius. Let :x;=linear velocity of the crank-pin 

center. Then 

Vh 20 ft. per minute 

20 ft. per minute X VI2 



or x = 



Vh 



X is therefore expressed in loiown terms. If now x, the space 
the crank-pin center is required to move through per minute, 
be divided by the space moved through per revolution, the result 
will equal the number of revolutions per minute =iV; 



N 



27rX length of crank* 



29. Lever-crank Quick Return — Fig. 23 shows a compound 
mechanism. The link d is the supporting frame or fixed link, 
and a rotates about ad in the direction indicated, communicating 



MOTION IN MECHANISMS. 



ZZ 



motion to c through the slider h so that c vibrates about cd. The 
link e, connected to c by a turning pair at ce, causes / to slide 
horizontally on another part of the frame or fixed link d. The 
center of the crank-pin, ah^ is given a constant linear velocity, 
and the slider, /, has motion toward the left with a certain mean 
velocity, and returns toward the right with a greater mean velocity. 
This is true because the slider / moves toward the left while a 
moves through the angle a ; and toward the right while a moves 
through the angle /?. But the motion of a is uniform, and hence 
the angular movement a represents more time than the angular 
movement /?; and /, therefore, has more time to move toward 
the left than it has to move through the same space toward the 
right. It therefore has a *' quick-return " motion. 



^ 



^/ 




nd« 



Fig. 23. 

The machine is driven so that the crank-pin center moves 
uniformly, and the velocity, at all points of its stroke, of the 
slider carrying a cutting-tool, is required. The problem, there- 
fore, is to find the relation of linear velocities of ef and ah for a 
series of positions during the cycle; and to draw the diagram 
of velocity of ej. 

Solution. — ah has a constant known linear velocity. The 
point in the link c which coincides, for the instant, with ah, re- 
ceives motion from ah , but the direction of its motion is different 



34 MACHINE DESIGN. 

from that of ab, because ab rotates about ad, while the coin- 
ciding point of c rotates about cd. If ab-A be laid off repre- 
senting the linear velocity of ab, then ab-B will represent the 
linear velocity of the coinciding point of the link c. Let the 
latter point be called x. 

Locate cf, at the intersection of e with the line cd-ad. Now 
cj and X are both points in the link c, and hence their linear 
velocities, relatively to the fixed link d, are proportional to their 
distances from cd. These two distances may be measured 
directly, and with the known value of linear velocity oix = ab-B 
give three known values of a simple proportion, from which the 
fourth term, the linear velocity of cj, may be found. 

Or, if the line BD be drawn parallel to cd-ad, the triangle 
B-D-ab is similar to the triangle cd-cf-ab, and from the simi- 
larity of these triangles it follows that BD represents the linear 
velocity of cf on the same scale that ab-B represents the linear 
velocity of x. Hence the linear velocity of cf, for the assumed 
position of the mechanism, becomes kno^\Ti. But since cf is a 
point of the slider, all of whose points have the same linear velocity 
because its motion relatively to d is rectilinear translation, it 
follows that the linear velocity of cf is the required linear velocity 
of the slider. At ef erect a line perpendicular to the direction 
of motion of the slider having a length equal to BD. 

This solution may be made for as many positions of the 
mechanism as are necessary to locate accurately the velocity 
curve. The ordinates of this curve will, of course, be the veloci- 
ties of the slider, and the abscissae the corresponding positions 
of the slider. 

Having drawn the velocity diagram, suppose that it is required 
to make the maximum linear velocity of the slider on the slow 
stroke =(2 feet per minute. Then the linear velocity of the 
crank-pin center ab=y can be determined from the propor- 
tion 



MOTION IN MECHANISMS. 

y vector A-ah 

Q maximum ordinate of velocity diagram* 
vector A-ab 



35 



y=Q 



maximum ordinate of velocity diagram' 



If r = the crank radius, the number of revolutions per minute = 



27:r 



A^Tien this mechanism is embodied in a machine, a becomes 
a crank attached to a shaft v^hose axis is at ad. The shaft turns 
in bearings provided in the machine frame. The crank carries a 
pin whose axis is at a6, and this pin turns in a bearing in the 
sliding block b. The link c becomes a lever keyed to a shaft 
whose axis is at cd. This lever has a long slot in which the block 
b slides. The link e becomes a connecting-rod, connected to both 
c and / by pin and bearing. The link / becomes the "cutter- 
bar " or "ram" of a shaper: the part which carries the cutting- 
tool. The link d becomes the frame of the machine, which not 
only affords support to the shafts at ad and cd, and the guiding 
surfaces for /, but also is so designed as to afford means for holding 
the pieces to be planed, and supports the feed mechanism. 




Fig. 24. 



30. Whitworth Quick Return. 

pound linkage, d is fixed, and 



-Fig. 24 shows another com- 
rotates uniformly about cJ, 



3^ MACHINE DESIGN. 

communicating an irregular rotary motion to a through the slider 
h. a is extended past ad (the part extended being in another 
parallel plane), and moves a slider / through the medium of 
a link e. This is called the " Whitworth quick-return mechanism." 
The point he, at which c communicates motion to a, moves along 
a, and hence the radius (measured from ad) of the point at which 
a receives a constant linear velocity varies, and the angular 
velocity of a must vary inversely. Hence the angular velocity 
of a is a maximum when the radius is a minimum, i.e., when 
a and c are vertical do^vnward; and the angular velocity of a 
is minimum when the radius is a maximum, i.e., when a and c 
are vertical upward. 

31. Problem. — To design a Whitworth Quick Return for a 

given ratio, 

mean VI of / forward 

mean VI of / returning* 

When the center of the crank-pin, C, reaches A, the point D will 

coincide with B, the link c will occupy the angular position cd-B, 

and the sHder / will be at its extreme position toward the left. 

When the point C reaches F, the point D will coincide with 
E, the link c will occupy the angular position cd-E, and the 
slider / will be at its extreme position toward the right. 

Obviously, while the link c moves over from the position 
cd-E to the position cd-B, the slider / will complete its forward 
stroke, i.e., from right to left. While c moves under from cd-B 
to cd-E, j will complete the return stroke, i.e., from left to right. 
The link c moves with a uniform angular velocity, and hence the 
mean velocity of / forward is inversely proportional to the angle 
^ (because the time consumed for the stroke is proportional to 
the angle moved through by the crank c), and the mean velocity 
of / returning is inversely proportional to a. Or 

mean VI of / forward a 
mean VI of / returning /?* 



MOTION IN MECHANISMS. 37 

For the design the distance cd-ad must be known. This may 
usually be decided on from the limiting sizes of the journals at cd 

and ad. Suppose that the above ratio =-5- = J, that cd-ad = 2! \ 

and that the maximum length of stroke of j=i2'\ Locate cd 
and measure off vertically downward a distance equal to 3'', 
thus locating ad. Draw a horizontal hne through ad. The 
point ej of the sHder / will move along this line. Since 

^=|, and a+/? = 36o°, 

.*. a = 150° and /? = 2io°. 

Lay off a from cd as a center, so that the vertical line through 
cd bisects it. Draw a circle through B with cd as a center, B 
being the point of intersection of the bounding line of a with a 
horizontal through ad. The length of the Hnk c = cd-B. 

The radius a^-C must equal the travel of /-^2=6'^ This 
radius is made adjustable, so that the length of stroke may be 
varied. The connecting-rod, e, may be made of any convenient 
length. 

32. Problem. — To draw the velocity diagram of the sHder 
/ of the Whitworth Quick Return. The point he, Fig. 25, as a 
point of c has a known constant hnear velocity relative to d, and 
its direction of motion is always at right angles to a hne joining 
it to cd. That point of the Hnk a which coincides in this posi- 
tion of the mechanism with he, receives motion from he, but its 
direction of motion relative to d is at right angles to the Hne he- 
ad. If he- A represents the linear velocity of he, its projection 
upon he-ad extended will represent the Hnear velocity of the 
point of a which coincides with he. Call this point x. Locate 
the centro a/, draw the line aj-he and extend it to meet the vertical 
dropped from B to C. The centro aj may be considered as a 



38 



MACHINE DESIGN. 



point in a, and its linear velocity relative to d^ when so considered, 
is proportional to its distance from ad. Hence 

VI of af _ ad-af 
VI of X ad-bc ' 




Fig. 25. 
But the triangles ad-af-bc and B-C-bc are similar. Hence 

VI of af BC 
Vloix~ B-bc 

This means that BC represents the linear velocity of aj upon the 
same scale that B-bc represents the linear velocity of x. But 
aj is a point in /, and all points in / have the same Hnear velocity 
relative to d since the motion is rectihnear translation; hence 
BC represents the hnear velocity of the sHder / for the given 
position of the mechanism, and it may be laid o£f as an ordinate 
of the velocity curve. This solution may be made for as many 
positions as are required to locate accurately the entire velocity 
curve for a cycle of the mechanism. 



CHAPTER III. 

PARALLEL OR STRAIGHT-LINE MOTIONS. 

33. Watt Parallel Motion. — Rectilinear motion in machines 
is usually obtained by means of prismatic guides. It is some- 
times necessar}^, however, to accompHsh the same result by 
linkages. 

The simplest and most widely known linkage used for giving 
rectilinear motion, to a point without the use of any sliding pairs 
is the so-called Watt Parallel Motion. It is one of the numerous 
inventions of James Watt and x 

consists of four links, three • ^ lab 

moving and one fixed, all con- 
nected by turning pairs, d, 
Fig. 26, is the fixed link, a 

rotates relative to d about ad; 

Fig. 26. 
c rotates relative to a about ca. 

The mechanism is shown in the position corresponding to the 

middle of its motion. 

As the points ab and be swing in the dotted arcs, the point P 

will travel in approximately a straight hne. The whole path of 

P is a lemniscate, but the part which is ordinarily used approaches 

very closely to a straight Hne. 

34. Parallelogram. — A true parallel motion is given by the 
Parallelogram which is shown in Fig. 27. The links a, b, c, and e 
are connected to each other by turning pairs, and the linkage is 
attached to the fixed link d by a. turning pair at ac. (This point 
is also ad and cd.) 

39 




40 



MACHINE DESIGN. 



The lengths ac — ah and ce — he are equal, as are also ac — ce and 
ah — he. The point P is fixed on e. Draw a line from P to ac ; it 
cuts the link h at P'. 

By similar triangles, 

F-he P-he 



ac—ce 



\P'-he=^ac- 



ce 



' P-ce 

/ P-he \ 

[p-cej 



= a constant. 



Therefore the point P' will lie at the same position on h for all 
positions of the mechanism. Likewise, by similar triangles, the 

P —ac P—^e 
ratio -z:: = -: ^^— = a constant for all positions of the 



ac 



he—ce 



mechanism. Since the line P—P' swings, relative to dj about the 



(XCifCLd^ 




Fig. 27 



pole ad (ac, cd) at every instant, it is obvious that the motions of 

P and P' relative to d will be similar to each other in every respect 

p 



and always in the ratio 



ac 
ac 



It follows that, if either of 



these points is guided to move in a straight-line path, the other 
point is constrained to move in a similar parallel path.* 

* The following demonstration is given for those who prefer an accurate 
proof : 

The position of the mechanism in Fig. 27 is taken as a perfectly general one 
and, in the same way, the instantaneous motion of the point P is assumed as 
indicated by the arrow. By methods indicated in earlier chapters locate the 
centres de and hd. Continue the lines ad — ah and he — ah until they cut the line 
P — de at X and y, respectively. It is necessary to prove that P' will always 



PARALLEL OR STRAIGHT-LINE MOTIONS. 41 

35. Grasshopper Motion. — A device which may be used to 
change the direction of rectihnear motion through a right angle 
is the linkage known as the Grasshopper Motion, shown in Fig. 28. 
This is the ordinary slider-crank chain with crank a and con- 
necting-rod h of equal length. The linkage is further modified 

move in a path parallel to P's motion and at a constant proportion to it. If 
this is true for instantaneous motion it is true for any motion P may be given 
relatively to d. Draw the line P' — hd. It can be shown that this line will 
always be parallel to P — de, for, since ac — x is parallel to ce—P 



Also, by similar triangles 
Hence, 



p- 


-be 


x- 


-bd 


be- 


— ce 


~bd- 


— ac' 


P- 


-be 


P- 


-P' 


be 


-ce 


P' 


—ac 


X- 


-bd 

= 


P- 


-P' 



bd—ac P' —ac 



which, considering the triangles ac—x — P and ac — bd—P\ shows that P' — bd 
is parallel to P—x, or to P — de. 

But P is a point of e and as such has an instantaneous motion relative to d in 
a direction perpendicular to P — de. In the same way, P' is a point of b and as 
such, relatively to d, has instantaneous motion perpendicular to P' — bd. These 
two instantaneous motions are therefore parallel. 

It remains to be shown that they will always be in the same proportion as to 

extent. The extent will be directly proportional to the instantaneous linear 

velocities. Both P and be as points of e rotate for the instant about de 

, . , , VIP P-de 

relatively to d, .•. — — -=- — . 

^ Mbe be-de 

Both be and P' are points of b, and as such, relatively to d, rotate about bd, 
Vibe be-bd be-de ,^ . , . , ^ 

Multiplying, 

VIP Vibe P-de be-de VIP P-de ,, . ., . , . 

VTb^^VlP^=-S^:^e^-^^^' "' ^.^ = p:^ = (by similar triangles). 

P-be 



be—ce 



■■ a constant value. 



Or, in other words, the linear velocities of P and P' bear a constant ratio to each 
other for all positions of the mechanism, and hence, these points will trace 
proportionately similar paths on d. — Q.E.D. 



42 



MACHINE DESIGN. 



by extending h beyond ah to a point P such that the length 
P-ah = crank length. It is obvious that P is constrained to move 
relative to </ in a straight-line path perpendicular to d through 
ad.^ 

36. General Method for Parallel-motion Design. — ^A general 
method of design which is applicable in many cases is as follows. 
In Fig, 29 J is the fixed link, and a is connected with it by a sliding 
pair, a, b, c, and e are connected by turning pairs, as shown. 
The constrainment is not complete because B is free to move 
in any direction, and its motion would, therefore, depend upon 
the force producing it. It is required that the point B shall move 



ce be c 



de 



p 

e h 

af 



^0 



M 






Fm. 29. 



in a straight line parallel to a. Suppose that B is caused to move 
along the required line; then any point of the hnk c, as ^, will 
describe some curve, FAE. If a pin be attached to c, with its 
axis at A, and a curved slot fitting the pin, with its sides parallel 
to FAEj be attached to d, as in Fig. 30, it follows that B can only 
move in the required straight line. This is the mechanism of the 
Tabor Steam-engine Indicator. 

* This is true because, from the construction of the mechanism, the line P-hd 
must always lie parallel to d. The point P, which rotates about hd as center rela- 
tively to d, always has an instantaneous motion perpendicular to P-hd and, con- 
sequently, perpendicular to d. 



PARALLEL OR STRAIGHT-LINE MOTIONS. 



43 



The curve described by A might approximate a circular arc 
whose center could be located, say, at O, Fig. 30. Then the 




Fig. 30. 

curved slot might be replaced by a link attached to d and c by 
turning pairs at O and A. This gives B approximately the 
required motion. This is the mechanism of the Thompson 
Steam-engine Indicator. 

If, while the point B is caused to move in the required straight 
line, a point in h, as P, Fig. 29, were chosen, it would be found 




Fig. 31. 

to describe a curve which would approximate a circular arc, 
whose center, O, and radius, =r, could be found. Let the link 



44 



MACHINE DESIGN 



whose length = r he attached to d and b by turning pairs whose 
axes are at O and P, and the motion of B will be approximately 
the required motion. This is the mechanism of the Crosby 
Steam-engine Indicator. One very important fact, however, 
is to be noted in connection with all steam-engine indicator 
pencil mechanisms. While it is important that the pencil point 



ac3 



cdi 







- ] ■_ ~— 






' y' 






y 






^ 












1 y 








^-■^v^ 




y 1 

y 1 




^y^ 


\^^>c. 



C2 



Parallel to 
\ a's motion 



df 



9A 



Line 



hci 



aci 



ah 



ah 



ah 



[S^icejand cdi 



ejf-ei e^ 



cd 



4 ad' 



Fig. 30a. 

B (Figs. 29 and 30) travel in a straight-line path parallel to the 
axis of the piston rod a, it is fully as important that the motion 
of the point B always be exactly the same multiple of a's motion. 
To determine in any given case whether this is true or not, lay 
off very accurately and to a large scale, say five times actual size, 
a skeleton outline of the mechanism for three positions. See 
Fig. ^oa. These positions are taken so that the total distance 



PARALLEL OR STRAIGHT-LINE MOTIONS. 45 

B^ — B^ represents the allowable range of the instrument as 
stated by the maker, usually about 3''. B^ is located at the mid- 
position. The links a, 6, c, e and /are drawn for each case in 
their proper relative positions, d being considered as the fixed 
link. 

The subscripts i, 2 and 3 refer to the positions of the links 
corresponding to the three pencil positions B^, B^ and By 

First take position B^. The centros df and de are located 
at once because they are permanent centers as well. Since a's 
motion relative to d is rectilinear translation, the centro ad will 
lie at infinity in a direction at a right angle to the direction of 
motion, or, in this case, at horizontal infinity. The centros c/^, 
a?>j, ftCj, and ce^ are located at once at the axes of the turning 
pairs connecting the respective links. 

Using Kennedy's theorem locate cd^ (on lines de—ce^ and 
df—cf^), and ac^ (on lines cd^ — ad and ah^ — hc^)^ At the 
instant in question, every point of c relatively to d is rotating 
about the centro cd^ and each point will have a linear velocity 
proportional to its distance from cd^. But B^ and ac^ are both 

pomts of c. Hence we may write J- = —^ -1 • But ac. 

VIB^ B^ —cd^ 

is also a point of a and at any instant every point of a has the 

same velocity relatively to d because the motion of a relative to d 

is rectilinear translation .*. -— — = —^ -^ • 

VIB^ B,-cdi 

Similarly for the second position, 

VI a ac^— cd^ 

-6 6 , 



VI B, B, - cd^ 



And for the third position, 

VI a _ ac^— cd 
VI B, ~ B,-cd^ 



3. 



46 MACHINE DESIGN. 

But, for proper action, 

Via Via Via . . r ^^ 

T.; p = T/7 p = T77 p "^ ^ constant for all positions, 

.*. rr* r should equal -r-^ --^, and also equal -rr ?, 

B^-cd^ ^ B2-cd^y ^ Bs-cd^ 

otherwise the diagrams will give a distortion of the piston, a's, 
motion. Also for true parallel motion B^^ should lie on the same 
horizontal through cd^ on which ac^ lies; B^ on the horizontal 
through cd^; and ^3 on the horizontal through cd^. 

An examination of existing indicator mechanisms in this 
manner gives very interesting results, and separates clearly 
those instruments which distort from those which are correct. 

37. Problem. — In Fig. 31 5 is the fixed axis of a counter- 
shaft; C is the axis of another shaft which is free to move in 
any direction. It is required to constrain D to move in the 
straight line EF. If D be moved along EF, sl tracing-point 
fixed at A in the link CD will describe an approximate circular 
arc, HAK, whose center may be found at O. A link whose 
length is OA may be connected to the fixed link, and to the link 
CD, by means of turning pairs at O and A. D will then be 
constrained to move approximately along EF. A curved slot 
and pin could be used, and the motion would be exact.* 

* Descriptions of many varieties of parallel motions may be found in Rankine's 
"Machinery and Millwork"; Weisbach's "Mechanics of Engineering," Vol. III^ 
"Mechanics of the Machinery of Transmission"; Kennedy's "Mechanics of 
Machinery"; and elsewhere. 



CHAPTER IV. 

CAMS. 

38. Cams Defined. — A machine part of irregular outline, as 
A, Fig. 32, may rotate or vibrate about an axis O, and commu- 
nicate motion by line contact to another machine part, B. A is 
called a cam. A cyHnder A, Fig. ^^, having a groove of any 
form in its surface, may rotate about its axis, CD, and communi- 
cate motion to another machine part, B. ^ is a cam. A disk 
A, Fig. 34, having a groove in its face, may rotate about its axis, 
O, and communicate motion to another machine part, B. A 
is a cam. In fact it is only a modification of A, Fig. 32. In 
designing cams it is customary to consider a number of simul- 
taneous positions of the driver and follov^er. The cam curve 
can usually be drawn from data thus obtained. 

39. Case I. — The follower is guided in a straight line, and 
the contact of the cam with the follower is always in this Hne. 
The line may be in any position relatively to the center of rota- 
tion of the cam; hence it is a general case. The point of the 
follower which bears on the cam is constrained to move in the 
line MN, Fig. 35. O is the center of rotation of the cam. About 
O as a center, draw a circle tangent to MN at /. Then A, B, 
C, etc., are points in the cam. When the point ^ is at / the 
point of the follower which bears on the cam must be at A^; 
when 5 is at / the follower- point must be at B'; and so on through 
an entire revolution. Through A, B, C, etc., draw lines tangent 

47 



48 



MACHINE DESIGN. 



to the circle. With O as a center, and OA' as a radius, draw a 
circular arc A'A'^, intersecting the tangent through A at A^'. 
Then A'^ will be a point in the cam curve. For, if A returns to 
J, A A'' will coincide with J A', A^^ will coincide with A^, and the 





Fig. 32. 



Fig. 2>Z' 





Fig. 34. 



Fig. r^, 



cam will hold the follower in the required position. The same 
process for the other positions locates other points of the cam 
curve. A smooth curve drawn through these points is the 
required cam outhne. Often, to reduce friction, a roller attached 
to the follower rests on the cam, motion being communicated 
through it. The curve found as above will be the path of the 
axis of the roller. The cam outline will then be a curve drawn 



CAMS. 49 

inside of, and parallel to, the path of the axis of the roller, at a 
distance from it equal to the roller's radius. Contact between 
the follower and the cam is not confined to the line MN if a 
roller is used. 

40. Case II. — The cam engages w^ith a surface of the follower, 
and this surface is guided so that all of its positions are parallel. 
The method given is due to Professor J. H. Barr. O, Fig. 36, 
is the center of rotation of the cam. The follower surface 
occupies the successive positions i, 2, 3, etc., when the Hnes 

A, B, C, etc., of the cam coincide with the vertical Hne through 
C. It is required to draw the outHne of a cam to produce the 
motion required. Produce the vertical line through O, cutting 
the positions of the follower surface in A^, B\ C ^ etc. With O 
as a center and radii OB' ^ 0C\ etc., draw arcs cutting the lines 

B, C, D, etc., in the points B'', C, D" , etc. Position i is the 
lowest position of the follower surface; therefore A must be in 
contact with the follower surface in the vertical Hne through O, 
because if the tangency be at any other point the motion in one 
direction or the other will lower the follower, which is not allow- 
able. A is therefore one point in the cam curve. Draw a 
line MN through B" at right angles to B"0, and rotate B"0 
till it coincides with B'O. Then the line MN will coincide with 
the position of the follower surface 2B' . But the cam curve 
must be tangent to this line when B coincides with B'O^ and 
therefore the line MN is a line to which the cam curve must be 
tangent. Similar lines may be drawn through the points C" ■, 
B" ^ etc. Each will be a Hne to which the cam curve must be 
tangent. Therefore, if a smooth curve be drawn tangent to ah 
these lines, it will be the required cam outline. 

41. Case III. — This is the same as Case II, except that the 
positions of the follower surface, instead of being parallel, con- 
verge to a point, 0\ Fig. 37, about which the foUower vibrates. 
The solution is the same as in Fig. 36, except that the angle be- 



so 



MACHINE DESIGN, 



tween the lines correspondnig to MN, Fig. 7,6, and the radial 
lines, instead of being a right angle, equals the angle between 
the corresponding position of the follower surface and the vertical. 




Fig. 36 



In these cases the cam drives the follower in only one direc- 
tion; the force of gravity, the expansive force of a spring, or 
some other force must hold it in contact with the cam. To 
drive the follower in both directions, the cam surface must be 
double, i.e., it takes the form of a groove engaging with a pin or 
roller attached to the follower, as in Fig. 34. 

This method is inclined to produce excessive wear. A better 
method is to have the follower provided with two rollers on op- 
posite sides of the cam-shaft. See Fig. 38. 

Cam A is designed to give the desired motion to the follower 
through the medium of roller i. Every position of this roller 
causes roller 2 to occupy a definite position, and the complementary 
cam B is so designed as to correspond to these positions of roller 
2. Cam B is rigidly mounted on the same shaft as ^, so that 
the two cams have no motion relative to each other. If the line 
of action of the follower passes through the center of the cam- 



CAMS. 



5^ 



shaft as shown in Fig. 38, it becomes a wtry simple matter to 
draw the outline of cam B\ all that is necessary is to keep the 






^^J 



J2r 




Fig. 38. 

sum of the radial lengths a + b=si constant == the distance be- 
tween the centers of rollers i and 2. 

42. Case IV. — To lay out a cam groove on the surface of a 
cylinder. — Aj Fig. 39, is a cylinder which is to rotate continu- 
ously about its axis. B can only move parallel to the axis of A. 
B may have a projecting roller to engage with a groove in the 
surface of A. CD is the axis of the roller in its mid-position. 
EF is the development of the surface of the cylinder. During 



52 



MACHINE DESIGN. 



the first quarter-revolution of A, CD is required to move one 
inch toward the right with a constant velocity. Lay off GH = i'\ 
and HJ = \KF^ locating J. Draw G/, which will be the middle 
line of the cam-groove. During the next half-revolution of A 
the roller is required to move two inches toward the left with a 
uniformly accelerated velocity. Lay off JL = 2'', and LM = ^KF. 
Divide LM into any number of equal parts, say four. Divide 
JL into four parts, so that each is greater than the preceding one 




Fig. 39. 

by an equal increment. This may be done as follows: i + 2-F 
3-1-4 = 10. Lay off from /, 0.1 JL, locating a; then 0.2JL from 
a, locating b; and so on. Through a, b, and c draw vertical 
lines; through m, n, and draw horizontal lines. The in- 
tersections locate d, e, and /. Through these points draw the 
curve from / to If, which will be the required middle line 
of the cam-groove. During the remaining quarter- re volution 
the roller is required to return to its starting-point with a 



c^MS 53 

uniformly accelerated velocity. The curve MN is drawn in 
the same way as JM. On each side of the line GJMN lay ofif 
parallel lines, their distance apart being equal to the diameter of 
the roller. Wrap EF upon the cylinder, and the required cam- 
groove is located. 



CHAPTER V. 

ENERGY IN MACHINES. 

43. The subject of motion and velocity, in certain simple 
machines, has been treated and illustrated. It remains now to 
consider the passage of energy through similar machines. From 
this the solution of force problems will follow. 

During the passage of energy through a machine, or chain of 
machines, any one, or all, of four changes may occur. 

I. The energy may be transferred in space. Example. — En- 
ergy is received at one end of a shaft and transferred to the other 
end, where it is received and utiHzed by a machine. 

II. The energy may be converted into another form. Exam- 
ples. — (a) Heat energy into mechanical energy by the steam- 
engine machine chain, {h) Mechanical energy into heat by fric- 
tion. U) Mechanical energy into electrical energy, as in a 
dynamo-electric machine; or electrical energy into mechanical 
energy in the electric motor, etc. 

III. Energy is the product of a force factor and a space factor. 
Energy per unit time, or rate oj doing work, is the product of a 
force factor and a velocity factor, since velocity is space per unit 
time. Either factor may be changed at the expense of the other; 
i.e., velocity may be changed, if accompanied by such a change 
of force that the energy per unit time remains constant. Cor- 
respondingly, force may be changed at the expense of velocity, 
energy per unit time being constant. Example. — A belt trans- 
mits 6000 foot-pounds per minute to a machine. The belt veloc- 
ity is 120 feet per minute, and the force exerted is 50 lbs. Fric- 

54 



ENERGY IN MACHINES. 55 

tional resistance is neglected. A cutting-tool in the machine 
does useful work; its velocity is 20 feet per minute, and the re- 
sistance to cutting is 300 lbs. Then, energy received per minute 
= 120X50 = 6000 foot-pounds; and energy delivered per minute 
= 20X300 = 6000 foot-pounds. The energy- received therefore 
equals the energy delivered. But the velocity and force factors 
are quite different in the two cases. 

IV. Energy may be transferred in time. In many machines 
the energy received at every instant equals that dehvered. There 
are many cases, however, where there is a periodical demand for 
work, i.e., a fluctuation in the rate of doing wcrk; while energy 
can only be supplied at the average rate. Or there may be a uni- 
form rate of doing work, and a fluctuating rate of supplying 
energy. In such cases means are provided in the machine, or 
chain of machines, for the storing oj energy, till it is needed. In 
other words, energy is transferred in time. Examples. — [a) In 
the steam-engine there is a var}'ing rate of supplying energy dur- 
ing each stroke, while there is (in general) a uniform rate of doing 
work. There is, therefore, a periodical excess and deficiency of 
effort. A heav}^ wheel on the main shaft absorbs the excess of 
energy with increased velocity, and gives it out again with re- 
duced velocity when the effort is deficient, (h) A pump dehvers 
water into a pipe system under pressure. The water is used in a 
hydrauhc press, whose action is periodic and beyond the capacity 
of the pump. A hydraulic accumulator is attached to the pipe 
system, and while the press is idle the pump slowly raises the 
accumulator weight, thereby storing potential energy, which is 
given out rapidly by the descending weight for a short time while 
the press acts, (c) A dynamo-electric machine is run by a steam- 
engine, and the electrical energy is delivered and stored in storage 
batteries, upon which there is a periodical demand. In this case, 
as well as in case (6), there is a transformation of energy as well 
as a transfer in time. 



56 MACHINE DESIGN, 

44. Force Problems. — Suppose the slider-crank mechanism in 
Fig. 40 to represent a shaping-machine, the velocity diagram of 




Fig. 40. 

the slider being drawn. The resistance offered to cutting metal 
during the forward stroke must be overcome. This resistance 
may be assumed constant. Throughout the cutting stroke there 
is a continually varying rate of doing work. This is because the 
rate of doing work = resisting force (constant) X velocity (vary- 
ing). This product is continually varying, and is a maximum 
when the slider's velocity is a maximum. The slider must be 
driven by means of energy transmitted through the crank a. The 
maximum rate at which energy must be supplied equals the maxi- 
mum rate of doing work at the slider. Draw the mechanism in 
the position of maximum velocity of slider; * i.e., locate the center 
of the slider-pin at the base of the maximum ordinate of the veloc- 
ity diagram, and draw b and a in their corresponding positions. 
The slider's known velocity is represented by y, and the crank- 
pin's required velocity is represented by a on the same scale. 
Hence the value of a becomes known by simple proportion. The 
rate of doing work must be the same at c and at ab (neglecting 
friction). t Hence Rvi= Fv2, in which R and vi represent the 

* It is customary to assume the slider's position for this condition to be that 
corresponding to an angle of 90° between crank and connecting-rod. This is not 
exactly true, but is a sufficiently close approximation for the ordinary proportions 
of crank and connecting-rod lengths. For method of exact determination of 
sHder's position see Appendix. 

t The effect of acceleration to redistribute energy is zero in this position, be- 
cause the acceleration of the slider at maximum velocity is zero, and the angular 
acceleration of b can only produce pressure in the journal at ad. If Ra equals 



ENERGY IN MACHINES. 57 

force and velocity factors at c; and F and v^ represent the tangen- 
tit*l force and velocity factors ab. R and v^ are known from the 
conditions of the problem, and V2 is found as above. Hence F may 

Rvi 
be found, = = force which, apphed tangentially to the crank- 
pin center, will overcome the maximum resistance of the machine. 
In all other positions of the cutting stroke the rate of doing work 
is less, and F would be less. But it is necessary to provide driv- 
ing mechanism capable of overcoming the maximum resistance, 
when no fly-wheel is used. If now F be multiplied by the crank 
radius, the product equals the maximum torsional moment ( = M) 
required to drive the machine. If the energy is received on some 
different radius, as in case of gear or belt transmission, the maxi- 
mum driving force = M -^ the new radius. During the return 
stroke the cutting-tool is idle, and it is only necessary to overcome 
the frictional resistance to motion of the bearing surfaces. Hence 
the return stroke is not considered in designing the driving mech- 
anism. When the method of driving this machine is decided on, 
the capacity of the driving mechanism must be such that it shall 
be capable of supplying to the crank-shaft the torsional driving 
moment M, determined as above. 

This method applies as well to the quick-return mechanisms 
given. In each, when the velocity diagram is dravm, the vector 
of the maximum linear velocity of the slider, =Li, and of the 
constant linear velocity of the crank-pin center, = L2, are kno^vn, 
and the velocities corresponding, vi and 7/2, are also known, from 
the scale of velocities. The rate of doing work at the slider and 

ii\e force necessary to produce acceleration of the slider mass at any position and 
Fa the force necessary at the crank pin to produce tangential acceleration of the 
rotating mass (assuming a variable velocity of the crank-pin as well as slider), 
;hen the equation in its most general form will be {R-\- Ra)vi= (F -\- Fa)v2. 
With uniform rate of rotation of the crank this becomes {R + Ra)Vi = Fv-2); and 
for position corresponding to maximum velocity of slider, as above, Rvi = Fi\. 



58 MACHINE DESIGN. 

at the crank-pin center is the same, friction being neglected. 
Hence Rvi =Fv2, or, since the vector lengths are proportional 

to the velocities they represent, RLi =FL2\ and F=—p — . There- 

fore the resistance to the slider's motion, =i?, on the cutting 

stroke, multiplied by the ratio of linear velocity vectors, 7-, of 

slider and crank-pin, equals F, the maximum force that must be 
applied tangentially at the crank-pin center to insure motion. 
F multiplied by the crank radius = maximum torsional driving 
moment required by the crank-shaft. If R is varying and known, 
find where Rv, the rate of doing work, is a maximum, and solve 
for that position in the same way as above. 

Where the mass to be accelerated is considerable the maxi- 
mum effort will be called for at the beginning of each stroke. 
If there is a quick return the maximum effort will come at the 
beginning of the return stroke. A planer calls for about twice 
as much power at the beginning of its return stroke as it does 
during its cutting stroke. 

45. Force Problems, Continued. — In the usual type of steam- 
engine the slider-crank mechanism is used, but energy is supplied 
to the slider (which represents piston, piston-rod, and cross- 
head), and the resistance opposes the rotation of the crank and 
attached shaft. In any position of the mechanism (Fig. 41), 
force applied to the crank-pin through the connecting-rod may 
be resolved into two components, one radial and one tangential. 
The tangential component tends to produce rotation; the radial 
component produces pressure between the surfaces of the shaft - 
journal and its bearing. The tangential component is approxi- 
mately a maximum when the angle between crank and connecting- 
rod equals 90°,* and it becomes zero when C reaches A or B, 
If there is a uniform resistance the rate of doing work is constant. 

* See foot-note on page 56. 



ENERGY IN MACHINES. 59 

Hence, since the energy is supplied at a varying rate, it follows 
that during part of the revolution the effort is greater than the 
resistance; while during the remaining portion of the revolution 
the effort is less than the resistance, and the machine will stop 
unless other means are provided to maintain motion. A *' fly- 
wheel " is keyed to the shaft, and this wheel, because of slight 




Fig. 41. 

variations of velocity, alternately stores and gives out the excess 
and deficiency of energy of the effort, thereby adapting it to the 
constant work to be done.* 

46. Problem. — Given length of stroke of the slider of a steam- 
engine slider-crank mechanism, the required horse-power, or 
rate of doing work, and number of revolutions. Required the 
total mean pressure that must be applied to the piston. 

Let L = length of stroke = i foot ; 
iJP = horse-power = 20; 
N = strokes per minute = 200 ; 
i^ = required mean force on piston. 
Then iV XL = 200 feet per minute = mean velocity of slider = F. 

Now, the mean rate of doing work in the cylinder and at the 
main shaft during each stroke is the same (friction neglected); 
hence i^F=iiZ'PX 33000, 

HP X 33000 20X33000 

F = f7 = =3300 lbs. 

V 200 "^^ 

* See Chapter XVI. 



6o 



MACHINE DESIGN. 



47. Solution of Force Problem in the Slider-crank Chain. 

— ^In the slider-crank chain the velocity of the slider necessarily 
varies from zero at the ends of its stroke to a maximum value 
near mid -stroke. The mass of the slider and attached parts 
is therefore positively and negatively accelerated each stroke. 
When a mass is positively accelerated it stores energy; and 
when it is negatively accelerated it gives out energy. The amount 
of this energy, stored or given out, depends upon the mass and 
the acceleration. The slider stores energy during the first part 
of its stroke and gives it out during the second part of its stroke. 



/fll 


V 




N 

K 
K ^ 

K 


m p\ 1 




Q 


^~~Mj ° 



Fig. 42. 

While, therefore, it gives out all the energy it receives, it gives 
it out differently distributed. In order to find exactly how the 
energy is distributed, it is necessary to find the acceleration 
throughout the slider's stroke. This may be done as follows: 
Fig. 42, A, shows the velocity diagram of the slider of a slider- 
crank mechanism for the forward stroke, the ordinates repre- 
senting velocities, the corresponding abscissas representing the 

, Jv 

slider positions. The acceleration required at any point =77, 

in which ^v is the increase in velocity during any interval of 



ENERGY IN My^ CHINES. 6 1 

time Jty assuming that the increase in velocity becomes constant 
at that point. Lay off the horizontal line OP=MN. Divide 
OP into as many equal parts as there are unequal parts in MN. 
These divisions may each represent Jt. At m erect the ordinate 
wn = Wi«i, and at o erect the ordinate op=Oipi. Continue 
this construction throughout OP, and draw a curve through the 
upper extremities of the ordinates. Fig. 42, B, is a velocity 
diagram on a ''time base." At O draw the tangent OT to the 
curve. If the increase in velocity were uniform during the time 
interval represented by Om, the increment of velocity would be 
represented by mT. Therefore mT is proportional to the accel- 
eration at the point O, and may be laid off as an ordinate of an 
acceleration diagram (Fig. 42C). Thus Qa = mT. The divi- 
sions of QR are the same as those of MN; i.e., they represent 
positions of the slider. This construction may be repeated for 
the other divisions of the curve B. Thus at n the tangent nTi 
and horizontal nq are dra^vn, and qTi is proportional to the 
acceleration at n, and is laid off as an ordinate be of the ac- 
celeration diagram. To find the value in acceleration units 
of Qa, mT is read off in velocity units =Jv by the scale 
of ordinates of the velocity diagram. This value is divided 
by Jt, the time increment corresponding to Om. The result 

Jv 
of this division -— = acceleration at M in acceleration units. 
Jt 

i/ = the time of one stroke, or of one half revolution of the crank 

divided by the number of divisions in OP. If the linear velocity 

of the center of the crank-pin in feet per second, =v, be repre- 

MN 
sented by the length of the crank radius = =a, then the scale 

of velocities, or velocity in feet per second for i inch of ordinate, 

V TzDN 
=— = — ^~. D is the actual diameter of the crank circle, N 
a a6o ' 

is the number of revolutions per minute, and a is the crank radius 

m.easured on the figure. 



62 MACHINE DESIGN, 

The determination of the acceleration curve, by means of 
tangents drawn to the ''time- base" velocity curve, has a serious 
drawback. The tangent lines are laid down by inspection, and slight 
inaccuracy in their location and construction may lead to consid- 
erable errors in the ordinates obtained for the acceleration curve. 

The following method is therefore suggested as an alternative. 

If one point is rotating about another point with a given 
instantaneous velocity =^' and a radius = r, the instantaneous 

radial acceleration of either point toward the other = —. 

Consider the slider-crank chain in the position at the begin- 
ning of the forward stroke as shown in Fig. 43^. The problem 
is to determine the acceleration of the point he toward ad. Ac- 
celerations toward the right will be considered as positive, toward 
the left as negative. In the position chosen the point ah is mov- 
ing, relatively to both links d and c, in the direction of the arrow 
with a velocity =1/, the uniform velocity of ah relatively to d. 

The acceleration of he toward ad is always made up of two 
components, namely, the acceleration of he toward ah and the 
acceleration of ah toward ad. In the position under considera- 

tion the acceleration of he toward ^^="r in a positive direction. 

Similarly the acceleration of ah toward ad=— in a positive di- 
rection. The total acceleration of he toward ad therefore equals 
the sum of these two components, or = -7- +— . 

On the other hand, at the end of the forward stroke, shown 
in Fig. 43^, the acceleration of he toward (^h = -r in a positive 

direction as before, while the acceleration of ah toward ad=— 

a 

in a negative direction. The algebraic sum of these two com- 



ENERGY IN MACHINES. 



63 



1-2 v^ 



ponents therefore = y - —. This quantity will always have a 

negative value, since in the slider- crank mechanism a must al- 
ways be smaller than h. 

To construct the acceleration curve, lay off a length MN 




Fig. 44. 



(Fig. 43Q proportionate to the length of the stroke of the slider. 

2 2 
At M erect an ordinate, MP, whose value equals t- + — It 

b a 

is best to use for these ordinates a scale on which a (the len^rth 



64 



MACHINE DESIGN. 



of the crank) represents the value — . At N erect the negative 



ordinate NQ=-r——.'^ 
^ b a 

There is a position of the slider, O, where the acceleration 

equals zero. This must correspond to the position of the slider 

* The following construction for graphically obtaining ordinates representing 
7~ + — and 7 , on the scale upon which a represents v 

I and, hence, a = — j is due to Professor Le Conte. 




Fig. 441. 



Reference is to Fig. 44a. M and N represent the position of the slider at the 
beginning and end of the stroke, respectively. 

At ab-^ erect the vector v{ = a) and from M draw a line through its upper extremity. 
Prolong this line until it cuts the perpendicular through ad, thus determining the 
length y^. 

At ab^ lay off downward the vector v { = a). From A^ draw a line to the lower 
extremity of this vector, cutting off' the length ^2 on the perpendicular tlu-ough ad 

y2 ,y2 ^2 ^2 

Then will y^ represent — H ; and j'g represent 7 • 



For, taking slider position M, by similar triangles, 



a + b 



b 



yi 



a2 -f ab 



— + a. But a = — , .■. y, 
b a 



r + - 

b a 



1' %/ ( = d^ 
For position A'', by similar triangles, -^-^ = — 7 — - 



72 = 



ab iP' 



= --a. 



ENERGY IN MACHINES. 6$ 

when it has its maximum velocity, which may be taken from 
the original velocity diagram of the slider, or, with greater 
accuracy, from Curve B in the Appendix. Through POQ draw 
a smooth curve- For most purposes this curve will be accurate 
enough. 

Where more points of the curve are desired for the sake of 
greater accuracy the method illustrated in Fig. 44 may be 
used. Assume the slider in the position at which its acceleration 
is desired and draw the crank a and connecting-rod b in their 
corresponding positions. Locate the centros ad, ah, ac, and bd. 
From ac draw a parallel to bc-ad until it cuts the crank, pro- 
longed if necessary, at A. From A draw a parallel to ad-ac 
until it cuts the connecting-rod at B. From B draw a perpen- 
dicular to the connecting-rod until it cuts bc-ad, prolonged if 
necessary, at C. Then ad-C is the desired ordinate of the 

acceleration diagram on the scale by which the length a==—. 

The proof is as follows, reference being made to Fig. 44. 

At this instant every point of b relatively to c is swinging 
about the centro be with a velocity proportional to its distance 
from be. 

vel. of bd rel. to c bd-bc ac-ad 
vel. of ab rel. to c ab-bc ac-ab' 

But ac-ad represents the velocity of c relatively to d (or d 
relatively to c) on the same scale that ad-ab represents the 
velocity of the point ab relatively to d. Therefore ac-ab repre- 
sents the velocity of ab rotating about be relatively to c on the same 
scale that ad-ab represents the velocity of ab relatively to d. 

Hence the radial acceleration of ab toward be * (or conversely 

* The acceleration of a point A with respect to another point B is the accel- 
eration of A with respect to a non-rotating body of which B is a point. 



66 MACHINE DESIGN. 



ah-ac 
of he toward ah) = — 7 — - , which is represented by the length 

ah-B, as can be shown as follows: 
By similar triangles 

ah-B ah-A ah-ac 



ah- 


■ac 


ah-ad 


ah-hc 










ah- 


-B- 


_ ah-ac 
ah-hc 


ah-ac 
~ b 










iXdX. 


ion 


of ah 


toward 


ad = 


ah- 

— — 


-a/ 


whose 



The radial ac 

a ' 

value we represent by the length a. The component of this 
acceleration in the direction hc-ah=ab-D. 

The acceleration of he, relatively to d, along the path hc-db 
is made up of two components: ist, the acceleration of he toward 
ah(=B-ah) plus, 2d, the acceleration of ah relatively to d along 
the same path {=ah-D). 

In the position shown this algebraic sum is the negative 
quantity represented by B-D. But the actual direction of tc's 
acceleration relatively to d is along the line hc-ad. Its accelera- 
tion in this direction must therefore be the quantity whose com- 
Donent along ah-hc is B-D, namely, C-ad. q.e.d. 

If the weight W of parts accelerated is known, the force F 
necessary to produce the acceleration at any slider position may 
be found from the fundamental formula of mechanics, 

p being the acceleration corresponding to the position considered. 
If the ordinates of the acceleration diagram are taken as repre- 
senting the forces which produce the acceleration, the diagram 
will have force ordinates and space abscissae, and areas wdll 
represent work. Thus, Qas, Fig. 42C, represents the work stored 



ENERGY IN MACHINES. 67 

during acceleration, and Rsd represents the work given out during 
retardation. Let MN, Fig. 45, represent the length of the 
slider's stroke and NC the resistance of cutting (uniform) on the 
same force scale as that by which Qa, Fig. 42C, represents the 

Wp 
force — - at the beginning of the stroke ; then energy to do cutting 

o 

per stroke is represented by the area MBCN, But during the 
early part of the stroke the reciprocating parts must be acceler- 
ated, and the force necessary at the beginning, found as above, 
= BD=Qa. The driving-gear must, therefore, be able to over- 
come resistance equal to MB + BD. The acceleration, and hence 
the accelerating force, decreases as the slider advances, becoming 
zero at E. From E on the acceleration becomes negative, and 
hence the slider gives out energy and helps to overcome the resist- 
ance, and the driving-gear has only to furnish energy represented 
by the area AEFN, though the work really 

-^- ' done against resistance equals that repre- 

^ sented by the area CEFN. The energy 



F represented by the difference of these areas. 

Fig. 45- =ACE, is that which is stored in the 

slider's mass during acceleration. Since by the law of con- 
servation of energy, energy given out per cycle = that received, 
it follows that area ^C£ = area DEB, and area BCMN == 
ADMN. This redistribution of energy would seem to modify 
the problem on page 52, since that problem is based on the 
assumption of uniform resistance during cutting stroke. The 
position of maximum velocity of slider, however, corresponds to 
acceleration =0. The maximum rate of doing work, and the 
corresponding torsional driving moment at the crank-shaft would 
probably correspond to the same position, and would not be 
materially changed. In such machines as shapers, the accelera- 
tion and weight of slider are so small that the redistribution of 
energy is unimportant. 



68 



MACHINE DESIGN. 



48. Solution of the Force Problem in the Steam-engine Slider- 
crank Mechanism. (Slider represents piston with its rod, and the 
cross-head.) — The steam acts upon the piston with a pressure 
which varies during the stroke. The pressure is redistributed 
before reaching the cross-head pin, because the reciprocating parts 
are accelerated in the first part of the stroke, with accompanying 
storing of energy and reduction of pressure on the cross-head 
pin; and retarded in the second part of the stroke, with accom- 
panying giving out of energy and increase of pressure on the 
cross-head pin. Let the ordinates of the full hne diagram above 
OX, Fig. 46^4, represent the total effective pressure on the piston 
throughout a stroke. Fig. 46B is the velocity diagram of slider. 




H::^ 



■="^ 




Fig. 46. 

Find the acceleration throughout stroke, and from this and the 
known value of weight of slider fmd the force due to acceleration. 
Draw diagram Fig. 46C, whose ordinates represent the force 
due to acceleration, upon the same force scale used in A. Lay 
off this diagram on OX as a base line, thereby locating the dotted 
line. The vertical ordinates between this dotted line and the 
upper line of A represent the pressure applied to the cross-head 
pin. These ordinates may be laid off from a horizontal base line, 
giving D. The product of the values of the corresponding ordinates 
of B and D =the rate oj doing ivork throughout the stroke. Thus 
the value of GH in pounds X value of EF in feet per second =the 



ENERGY IN MACHINES. 69 

rate of doing work in foot-pounds per second upon the cross- 
head pin, when the center of the cross-head pin is at E. The 
race of doing work at the crank-pin is the same as at the cross- 
head pin. Hence dividing this rate of doing work, ^EFxGH, 
by the constant tangential velocity of the crank-pin center, gives 
the force acting tangentially on the crank-pin to produce rotation. 
The tangential forces acting throughout a half revolution of 
the crank may be thus found, and plotted upon a horizontal 
base line=length of half the crank circle (Fig. 4jB). The work 
done upon the piston, cross-head pin, and crank during a piston 
stroke is the same. Hence the areas of A and D, Fig. 46, are 
equal to each other, and to the area of the diagram. Fig. 47^. 
The forces acting along the connecting-rod for all positions 
during the piston stroke may be found by drawing force triangles 
with one side horizontal, one vertical, and one parallel to position 
of connecting-rod axis, the horizontal side being equal to the 
corresponding ordinate of Fig. 46D. The vertical sides of these 
triangles will represent the guide reaction, while the side parallel 
to the connecting-rod axis represents the force transmitted by 
the connecting-rod. 

The tangential forces acting on the crank-pin may be found 
graphically by the method shown in Fig. 47^. Let GH repre- 
sent the net effective force acting in a horizontal direction at the 
center of the cross-head pin. 

It has been shown that EF represents the velocity of the 
slider on the same scale that EA represents that of the center of 
the crank-pin; also that the rate of doing work, after having 
made the necessary corrections for acceleration, is the same at 
the center of the crank-pin as at the slider, i.e., Gil X£i^= tan- 
gential force at center of crank-pin XEA. Hence the tangential 

EF 
force at center of crank-pin =G^XTr^. 



70 



MACHINE DESIGN. 



Lay off AB=GH, and draw BC parallel to EF. Then, by 
similar triangles, 

BC EF . ^^ J j^EF ^ttEF 

the tangential force acting at the crank-pin center for the assumed 
position of the mechanism, on the same scale as GiJ= net effective 
horizontal force on slider. 

Lay off AD=BC. 

Following through this construction for a number of positions 
of the mechanisms, a polar diagram is determined which shows 




Fig. 47B. 

very clearly the relation existing between the varying tangential 
forces and the corresponding crank positions. Before this dia- 
gram may be used in the solution of the fly-wheel problem (see 
Chapter XVI) it should be transferred to a straight-line base 
whose length for one stroke equals the semi-circumference of the 
crank-pin circle. That is, the abscissae will be the distance moved 
through by the center of the crank-pin and the ordinates will be 
the corresponding radial intercepts AD. The diagram so ob- 
tained will be identical with that shown m Fig. 47-B. 



CHAPTER VI. 

PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 

49. The size and form of machine parts * are governed by 
six main considerations: 

(i) The size and nature of the work to be accommodated (as 
the swing of engine-lathes, etc.)- 

(2) The stresses which they have to endure. 

(3) The maintaining of truth and accuracy against wear, in- 
cluding all questions of lubrication. 

(4) The cost of production. 

(5) Appearance. 

(6) Properties of materials to be used. 

The first is a given condition in any problem ; the second will 
be discussed here; the third will be treated in the chapters on 
Journals and Sliding Surfaces; the fourth is touched upon here; 
the principles governing the fifth are treated in Chapter XIX 
and here. 

It is assumed in this and following chapters that the reader is 
familiar with the properties of the materials employed in machine 
construction, f and with the general principles of the science of 
mechanics. 

50. The stresses acting on machine parts may be constant, 
variable, or suddenly applied. 



* On this general subject see an excellent article by Prof. Sweet in the Jour- 
nal of the Franklin Institute, 3d Series, Vol. 125, pp. 278-300. The reader is 
also referred to "A Manual of Machine Construction," by Mr. John Richards, 
and to the Introduction of this volume. 

t See Smith's "Materials of Machines." 

71 



72 MACHINE DESIGN. 

A CONSTANT stress is frequently spoken of as a steady, or 

DEAD, LOAD. 

A VARIABLE stress is known as a live load. 

A suddenly applied stress is known as a shock. 

51. Constant Stress. — If a machine part is subjected to a con- 
stant stress, i.e., an unvarying load constantly applied, its design 
becomes a simple matter, as the amount of such a stress can 
generally be very closely estimated. Knowing this and the 
properties of the materials to be used, it is only necessary to cal- 
culate the area which will sustain the load without excessive 
deformation. 

Thus, in simple tension or compression, if we let £/ = the ulti- 
mate strength of the material in pounds per square inch, F the 
total constant stress in pounds, A the unknown area in square 
inches necessarv to sustain F, we write 



A = 



U^K' 



where X is a so-called factor of safety, introduced to reduce 
the permitted unit stress to such a point as will limit the deforma- 
tion (strain) to an allowable amount, and also to provide for pos- 
sible defects in the material itself. In exceptional cases where 
the stresses permit of accurate calculation, and the material is of 
proven high grade and positively knovm strength, K has been 
given as low a value as ij; but values of 2 and 3 are ordinarily 
used for wrought iron and steel free from welds; while 4 to 5 are 
as small as should be used for cast iron, on account of the uncer- 
tainty of its composition, the danger of sponginess of structure, 
and indeterminate shrinkage stresses. 

The safe unit stress =/=-^ in pounds per square inch. 

52. Variable Stress. — ^We pass next to the consideration of 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 73 

variable stresses or live loads. Here the problem is much more 
complex than with dead loads. 

Experiments by Wohler,* and Bauschinger,f with the work 
of Weyrauch t and others have given us the laws of bodies sub- 
jected to repeated stresses. In substance Wohler's law is as 
follows: Material may be broken by repeated applications 

OF A FORCE WHICH WOULD BE INSUFFICIENT TO PRODUCE RUPTURE 
BY A SINGLE APPLICATION. ThE BREAKING IS A FUNCTION OF 
RANGE OF STRESS ; AND AS THE VALUE OF THE RECURRING STRESS 
INCREASES, THE RANGE NECESSARY TO PRODUCE RUPTURE DE- 
CREASES. If the stress be reversed, the range equals 

THE sum OF THE POSITIVE AND NEGATIVE STRESS. 

Bauschinger's conclusions were as follows: 
(i) With repeated tensile stresses whose lower limit 
WAS zero, and whose upper limit was near the original 

ELASTIC LIMIT, RUPTURE DID NOT OCCUR WITH FROM 5 TO 1 6 

MILLION REPETITIONS. He cautions the designer (a) that this 
will not hold for defective material, i.e., a factor of safety must 
still be used for this reason; and (b) that the elastic limit of the 
material must be carefully determined, because it may have been 
artificially raised by cold working, in which case it does not accur- 
ately represent the material. The original elastic limit may be de- 
termined by testing a piece of the material after careful annealing. 
(2) With often-repeated stresses varying between zero 
and an upper stress which is in the neighborhood of or 
above the elastic limit, the latter is raised even above, 
often far above, the upper limit of stress, and it is raised 
higher as the number of repetitions of stress increases, 

* "Ueberdie Festigkeitsversuche mit Eisen und Stahl," A. Wohler, Berlin, 1870. 

t "Mittheilungen der Konig. Tech. Hochschule zu Munchen," J, Bau- 
schinger, Munich, 1886 and 1897. 

X "Structures of Iron and Steel," by J. Weyrauch. Trans, by A. J. DuBois, 
New York, 1877 



74 M/ICHINE DESIGN, 

WITHOUT, HOWEVER, A KNOWN LIMITING VALUE, L, BEING EX- 
CEEDED. 

(3) Repeated stresses between zero and an upper limit 
below l do not cause rupture; but if the upper limit is 
above L rupture will occur after a limiting number of 
repetitions. 

From this it will be seen that keeping within the original 
ELASTIC LIMIT insures safety against rupture from repeated 
stress if the stress is not reversed; and that, when the stress is 
reversed, the total range should not exceed the original elas- 
tic RANGE of the material. 

Various formulae have been proposed by different authorities 
embodying the foregoing laws. 

Unwin's is here given as being most simple and general: 

Let U be the breaking strength of the material in pounds 
per square inch for a load once gradually applied. 

Let / max. be the breaking strength in pounds per square inch 
for the same material subjected to a variable load ranging be- 
tween the limits /max. and /min., and repeated an indefinitely 
great number of times, /min. is + if the stress is of the same 
kind as / max., and is — if the stress is of the opposite kind, 
and it is supposed that / min. is not greater than / max. Then 
the range of stress is J=/ max.=F/min., the upper sign being 
taken if the stresses are of the same kind and the lower if they 
are different. Hence J is always positive. The formula * is 



fmsix.= — +VU^-y}JU, 



where r; is a variable coefficient whose value has been experi- 
mentally determined. For ductile iron and steel ^ = 1.5, in- 
creasing with hardness of the material to a value of 2. 

* Unwin's "Machine Design," Vol. i, 1903, pp. 32-36. 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 75 

This formula is of general application. 
Three cases may be considered: 

(i) A constant stress, or dead load. In this case the range 
of stress, J = o, and consequently / max. = t/, as it should be. 

(2) The stress is variable between an upper limit and zero, 
but is not reversed. 

Here J = / max., since /min.=o, and consequently / max. = 

(3) The stress is reversed, being alternately a compressive 
and tensile stress of the same magnitude. 

Here / min. = — / max. and i = 2/ max. 



.*. /max. = — U. 

21) 



In each case it is necessary to divide the breaking load, / max., 

by a factor of safety in order to get the safe unit stress /, i.e,, 

f max 
f = - — —rr^. i^ is a factor of safety whose numerical value depends 

upon the material used. (See Sec. 51.) 

53. Problem. — Consider that there are three pieces to be 
designed using machinery steel having an ultimate tensile strength 
of 60,000 lbs. per square inch. 

The first piece sustains a steady load having a dead weight 
suspended from it. 

The second piece is a member of a ■ structure which is alter- 
nately loaded and unloaded without shock. 

The third piece is subjected to alternate stresses without 
shock. 

In each case the maximum load is the same, being 30,000 lbs. 
= F, This material is generally reliable and uniform in quality. 
A factor of safety of 3 is common; .'. K^^ in each case; 7^ = 1.5. 



7 6 MACHINE DESIGN. 

Case I. 

t TT ^ i /max. 
/ max. = U and / = , 



U 60000 
•'• j=~^ = 20,000 lbs. per sq. in. 

The necessary area A to sustain F is determined by the equation 
F 30000 ^ 

A =-r "= ^ = 1 1 SO. in. 

J 20C00 "^ ^ 

Case II. 



/ max. = 2 (V. 7^2-+ i — 7j)U. 
ry = 1.5 and C/ = 60,000 lbs. per sq. in. 
.*. / max. =.6o54Z7 = 36,324 lbs. per sq. in. 

. / max. ^ ,^ . 

/= = 12,108 lbs. per sq. m. 



F 30000 
A =-7 = Q ""^2 •^?- "^^-j nearly. 



Case III. 



/ max. = — Z7. 



60000 

/ max. = = 20,000 lbs. per sq. in. 



/ = =6667 lbs. per sq. in. 



. F 30000 

The importance of considering the question of range of stress 
in designing is brought home by this illustration. Comparison 



PROPORTIONS OF MACHINE P/iRTS AS DICTATED BY STRESS. 77 

of results shows that with the same maximum load in each 
case, the second piece must be given nearly twice the area of the 
first, while the third must be three times as great in area as the 
first, the only difterence in the three cases being the range of 
stress. 

54. Shock. — Consideration of the design of parts subjected 
to shocks or suddenly applied loads. 

(i) A load is applied on an unstrained member in a single 
instant, but without velocity. 

In this case, if the stress does not exceed the limit of elasticity 
of the material, the stress produced will be just twice that pro- 
duced by a gradually applied load of the same magnitude. If 
F= maximum total load as before, then the maximum total 
stress = 2F. The design of the member is then made as in Case 
II or Case III of the preceding section. 

(2) A load IS applied on an unstrained member in a single 
instant, but with velocity. 

In this case the stress on the member will exceed that due to 
a gradually applied load of the same magnitude by an amount 
depending on the energy possessed by the load at the moment 
of impingement. 

Assume that a member is stressed by a load F falling through 
a height h. The unknown area of the member =^, and the 
allowable strain {i.e., extension or compression) = A. x\s before, 
/= allowable unit stress (determined by the question of range by 
the use of Un win's formula). 

The energy of the falling load is 

F{h^-X). 

The work done in straining the member an amount X with a 
maximum fiber stress / is -XA, provided the elastic limit is 



78 MACHINE DESIGN. 

not exceeded. Equating these values 
work done and solving for A gives A 



not exceeded. Equating these values of energy expended and 

2F{h + X) 



55. Form Dictated by Stress. Tension. — Suppose that A and 
B, Fig. 48, are two surfaces in a machine to be joined by a member 
subjected to simple tension. What is the proper 
form for the member? The stress in all sec- 
^ tions of the member at right angles to the line 



"Ptp a S 

of appHcation,^5, of the force will be the same. 
Therefore the areas of all such sections should be equal; hence 
the outlines of the members should be straight lines parallel to 
AB. The distance of the material from the axis AB has no 
effect on its ability to resist tension. Therefore there is nothing 
in the character of the stress that indicates the form of the 
cross-section of the member. The form most cheaply produced, 
both in the rolling-mill and the machine-shop, is the cylindrical 
form. Economy, therefore, dictates the circular cross-section. 
After the required area necessary for safely resisting the stress is 
determined, it is only necessary to find the corresponding diam- 
eter, and it will be the diameter of all sections of the required 
member if they are made circular. Sometimes in order to get a 
more harmonious design, it is necessary to make the tension 
member just considered of rectangular cross-section, and this is 
allowable although it ahnost always costs more. The thin, wide 
rectangular section should be avoided, however, because of the 
difficulty of insuring a uniform distribution of stress. A unit 
stress might result from this at one edge greater than the strength 
of the material, and the piece would yield by tearing, although 
the AVERAGE stress might not have exceeded a safe value. 

56. Compression. — If the stress be compression instead of 
tension, the same considerations dictate its form as long as it 
is a "short block," i.e., as long as the ratio of length to lateral 
dimensions is such that it is sure to yield by crushing instead of 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 79 

by ''buckling." A short block, therefore, should have its longi- 
tudinal outlines parallel to its axis, and its cross-section may be 
of any form that economy or appearance may dictate. Care 
should be taken, however, that the least lateral dimension of 
the member be not made so small that it is thereby converted 
into a ''long column." 

If the ratio of longitudinal to lateral dimensions is such that 
the member becomes a "long column," the conditions that dic- 
tate the form are changed, because it would yield by buckling or 
flexure instead of crushing. The strength and stiffness of a 
long column are proportional to the moment of inertia of the 
cross-section about a gravity axis at right angles to the plane 
in which the flexure occurs. A. long column with "fixed" or 
"rounded" ends has a tendency to yield by buckling which is 
equal in all directions. Therefore the moment of inertia needs 
to be the same about all gravity axes, and this of course points to 
a circular section. Also the moment of inertia should be as large 
as possible for a given weight of material, and this points to the 
hollow section. The disposition of the metal in a circular hollow 
section is the most economical one for long-column machine 
members with fixed or rounded ends. This form, like that for 
tension, may be changed to the rectangular hollow section if 
appearance requires such change. If the long-column machine 
member be "pin connected," the tendency to buckle is greatest 
in a plane through the line of direction of the compressive force 
and at right angles to the axis of the pins. The moment of iner- 
tia of the cross-section should therefore be greatest about a gravity 
axis parallel to the axis of the pins. Example, a steam-engine 
connecting-rod. 

57. Flexure. — ^When the machine member is subjected to 
transverse stress the best form of cross-section is probably the I 
section, a. Fig. 49, in which a relatively large moment of inertia, 
with economy of material, is obtained by putting the excess of 



So MACHINE DESIGN. 

the material where it is most effective to resist flexure, i.e., at the 
greatest distance from the given gravity axis. Sometimes, how- 
ever, if the I section has to be produced by cutting away the 
material at e and d, in the machine-shop, instead of producing 
the form directly in the rolls, it is cheaper to use the solid rect- 
angular section c. If the member subjected to transverse stress 
is for any reason made of cast material, as is often the case, the 
form b is frequently preferable for the following reasons: 

(i) The best material is almost sure to be in the thinnest part 
of a casting, and therefore in this case at / and g, where it is most 
effective to resist flexure. 

(2) The pattern for the form b is more cheaply produced and 
maintained than that for a. The hollow box section, when 
permitted by considerations of construction and expense is still 
better. 

(3) If the surface is left without finishing from the mold, any 
imperfections due to the foundry work are more easily corrected 
in b than in a. 

Machine members subjected to transverse stress, which con- 
tinually change their p»osition relatively to the force which pro- 
duces the flexure, should have the same moment of inertia about 
all gravity axes, as, for instance, rotating shafts that are strained 
transversely by the force due to the weight of a fly-wheel, or that 
due to the tension of a driving-belt. The best form of cross- 
section in this case is circular. The hollow section would give 
the greatest economy of material, but hollow members are ex- 
pensive to produce in wrought material, such as is almost inva- 
riably used for shafts. The hollow circular section is meeting 
with increasing use, especially for large shafts, on account of the 
combined lightness and strength. 

58. Torsion. — Torsional strength and stiffness are propor- 
tional to the polar moment of inertia of the cross-section of the 
member. This is equal to the sum of the moments of inertia 
about two gravity axes at right angles to each other. The forms 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 8r 

in Fig. 49 are therefore not correct forms for the resistance of 
torsion. The circular solid or hollow section, or the rect- 
angular solid or hollow section, should be used. 

The I section, Fig. 50, is a correct form for resisting the stress 
P, applied as shown. Suppose the web C to be divided on the 
line CD, and the parts to be moved out so that they occupy 
the positions shown at a and b. The form thus obtained is 
called a "box section." By making this change the moment 
of inertia about ab has not been changed, and therefore the 
new form is just as effective to resist flexure due to the force P 
as it was before the change. The box section is better able to 



^^^Z2 







\'9 



p 

1 



J 



■^IG. 49- 



Fig. 50. 



resist torsional stress, because the change made to convert tne 
I section into the box section has increased the polar moment 
of inertia. The two forms are equally good to resist tensile and 
compressive force if they are sections of short blocks. But if 
they are both sections of long columns, the box section would be 
preferable, because the moments of inertia would be more nearly 
the same about all gravity axes. 

59. Machine Frames. — The framing of machines almost always 
sustains combined stresses, and if the combination of stresses 
include torsion, flexure in different planes, or long-column com- 
pression, the box section is the best form. In fact, the box sec- 
tion is by far the best form for the resisting of stress in machine 
frames. There are other reasons, too, besides the resisting of 
stress that favor its use.* 



* See Richard's "Manual of Machine Construction.' 



82 MACHINE DESIGN. 

(i) Its appearance is far finer, giving an idea of complete- 
ness that is always wanting in the ribbed frames. 

(2) The faces of a box frame are always available for the 
attachment of auxiliary parts without interfering with the per- 
fection of the design. 

(3) The strength can always be increased by decreasing the 
size of the core, without changing the external appearance of 
the frame, and therefore without any work whatever on the 
pattern itself. 

The cost of patterns for the two forms is probably not very 
different, the pattern itself being more expensive in the ribbed 
form, and the necessary core-boxes adding to the expense in the 
case of the box form. The expense of production in the foimdry, 
however, is greater for the box form than for the ribbed form, 
because core work is more expensive than ''green-sand" work. 
The balance of advantage is very greatly in favor of box forms, 
and this is now recognized in the practice of the best designers 
of machinery. 

To illustrate the appHcation of the box form to machine 
members, let the table of a planer be considered. The cross- 
section is almost universally of the form shown in Fig. 51. This 
is evidently a form that would yield easily to a force tending to 



Fig. 51. 

twist it, or to a force acting in a vertical plane tending to bend it. 
Such forces may be brought upon it by "strapping down work," 
or by the support of heavy pieces upon centers. Thus in Fig. 52 
the heavy piece E is supported between the centers. For proper 
support the centers need to be screwed in with a considerable 
force. This causes a reaction tending to separate the centers and 
to bend the table between C and D. As a result of this the V's 



PROPORTIONS OF MACHINE PARTS AS DICTATED BY STRESS. 83 

on the table no longer have a bearing throughout the entire 
surface of the guides on the bed, but only touch near the ends, 
the pressure is concentrated upon small surfaces, the lubricant is 
squeezed out, the V's and guides are "cut," and the planer is 



Fig. 52. 





Fig. 53. 



Fig. 54. 



rendered incapable of doing accurate work. If a table were made 
of the box form shown in Fig. 53, with partitions at intervals 
throughout its length, it would be far more capable of maintain- 
ing its accuracy of form under all kinds of stress, and would be 
more satisfactory for the purpose for which it is designed.* 

The bed of a planer is usually in the form shown in section 
in Fig. 54, the side members being connected by "cross-girts " 
at intervals. This is evidently not the best form to resist flexure 
and torsion, and a planer-bed may be subjected to both, either 
by reason of improper support or because of changes in the form 
of foundation. If the bed were of box section with cross parti- 
tions, it would sustain greater stress without undue yielding. 
Holes could be left in the top and bottom to admit of supporting 
the core in the mold, to serve for the removal of the core sand, 
and to render accessible the gearing and other mechanism inside 
of the bed. 



* Professor Sweet has designed and constructed such a table for a large mill- 
ing-machine. 



84 



MACHINE DESIGN. 



This same reasoning applies to lathe -beds. They are strained 
transversely by force tending to separate the centers, as in the 
case of '' chucking"; torsionally by the reaction of a tool cutting 
the surface of a piece of large diameter; and both torsion and 
flexure may result, as in the case of the planer-bed, from an 
improperly designed or yielding foundation. The box form 
would be the best possible form for a lathe-bed; some diffi- 
culties in adaptation, however, have prevented its extended use 
as yet. 

These examples illustrate principles that are of very broad 
application in the designing of machines. 

60. Brackets. — Often in machines there is a part that pro- 
jects either vertically or horizontally and sustains a transverse 





Fig. 55. 



stress; it is a cantilever, in fact. If only transverse stress is sus- 
tained, and the thickness is uniform, the outline for economy of 
material is parabolic. In such a case, however, the outline curve 
of the member should start from the point of application of the 
force, and not from the extreme end of the member, as in the 
latter case there would be an excess of material. Thus in A, 
Fig. 55, P is the extreme position at which the force can be ap- 



PROPORTIONS OF M/1 CHINE PARTS AS DICTATED BY STRESS. 85 

plied. The parabolic curve a is drawn from the point of appli- 
cation of P. The end of the member is supported by the auxil- 
iary' curve c. The curve h drawn from the end gives an excess 
of material. The curves a and c may be replaced by a single 
continuous curve as in C, or a tangent may be drawn to a at its 
middle point as in B, and this straight line used for the outline, 
the excess of material being shght in both cases. Most of the 
machine members of this kind, however, are subjected also to 
other stresses. Thus the ''housings" of planers have to resist 
torsion and side flexure. They are very often supported by two 
members of parabolic outline ; and to insure the resistance of the 
torsion and side flexure, these two members are connected at their 
parabolic edges by a web of metal that really converts them into 
a box form. Machine members of this kind may also be sup- 
ported by a brace, as in Z>. The brace is a compression member 
and may be stiffened against buckhng by a "web " as shown, or 
by an auxiliary brace. 

61. Other Considerations Governing Form. — One considera- 
tion governing the form of machine parts has been touched upon 
in the preceding sections. It may be well to state it here as a 
general principle: Other considerations being equal the form of 
a member should be that which can be most cheaply produced 
both as regards economy of material and labor. 

Another element enters into the form of cast members. Cast- 
' ings, unless of the most simple form, are ahnost invariably sub- 
jected to indeterminate shrinkage stresses. Some of these are 
undoubtedly due to faulty work on the part of the molder, others 
are induced by the very form which is given the piece by the de- 
signer. They cannot be eliminated entirely, but the danger can 
be mmimized by paying attention to these general laws: 

(a) Avoid all sharp corners and re-entrant angles. 

(b) All parts of all cross-sections of the member should be 
as nearly of the same thickness as possible. 



86 MACHINE DESIGN. 

(c) If it is necessary to have thick and thin parts in the same 
casting, the change of form from one to the other should be as 
gradual as possible. 

(d) Castings should be made as thin as is consistent with con- 
siderations of strength, stiffness, and resistance to vibration. 



CHAPTER VII. 

RIVETED JOINTS. 

62. Methods of Riveting. — A rivet is a fastening used to unite 
metal plates or rolled structural forms, as in boilers, tanks, built- 
up machine frames, etc. It consists of a head. A, Fig. 56, and a 
straight shank, B. It is inserted, usually red-hot, into holes, 
either drilled or punched in the parts to be connected, and the 
projecting end of the shank is then formed into a head (see dotted 
lines) either by hand- or machine-riveting. A rivet is a permanent 
fastening and can only be removed by cutting off the head. A 
row of rivets joining two members is called a riveted joint or 
SEAM OF RIVETS. In hand-riveting the projecting end of the 
shank is struck a quick succession of blows with hand hammers 
and formed into a head by the workman. A helper holds a sledge 




Fig. 56. 





Fig. 58. 



or ''dolly bar" against the head of the rivet. In ''button-set" 
or "snap" riveting, the rivet is struck a few heavy blows with a 
sledge to "upset" it. Then a die or "button set," Fig. 57, is 
held with the spherical depression, B, upon the rivet; the head, 
A, is struck with the sledge, and the rivet head is thus formed. 
In machine-riveting a die similar to B is held firmly in the ma- 
chine and a similar die opposite to it is attached to the piston of 

87 



SS MACHINE DESIGN. 

a steam, hydraulic, or pneumatic cylinder. A rivet, properly 
placed in holes in the members to be connected, is put between 
the dies and pressure is applied to the piston. The movable die 
is forced forward and a head formed on the rivet. 

The relative merits of machine- and hand- riveting have been 
much discussed. Either method carefully carried out will pro- 
duce a good serviceable joint. If in hand-riveting the first few 
blows be light the rivet will not be properly upset, the shank will 
be loose in the hole, and a leaky rivet results. If in machine- 
riveting the axis of the rivet does not coincide with the axis of 
the dies, an off-set head results. (See Fig. 58.) In large shops 
where work must be turned out economically in large quantities, 
machines must be used. But there are always places inacces- 
sible to machines, where the rivets must be driven by hand.* 

63. Perforation of Plates. — Holes for the reception of rivets 
are usually punched, although for thick plates and very careful 
work they are sometimes drilled. If a row of holes be punched 
in a plate, and a similar row as to size and spacing be drilled in 
the same plate, testing to rupture will show that the punched plate 
is weaker than the drilled one. If the punched plate had been 
annealed it would have been nearly restored to the strength of 
the drilled one. If the holes had been punched ^ inch to | inch 
small in diameter and reamed to size, the plate would have been 
as strong as the drilled one. These facts, which have been ex- 
perimentally determined, point to the following conclusions: 
First, punching injures the material and produces weakness. 
Second, the injury is due to stresses caused by the severe action 
of the punch, since annealing, which furnishes opportunity for 
equalizing of stress, restores the strength. Third, the injury 
is only in the immediate vicinity of the punched hole, since ream- 
ing out l^ inch or less on a side removes all the injured material. 

* See Sec. 75 for discussion of the importance of holding rivet under pressuie 
until it is cooled; and the advantage of large rivets over small. 



RiyETED JOINTS. 



89 



In ordinary boiler work the plates are simply punched and riveted. 
If better work is required the plates must be drilled, or punched 
small and reamed, or punched and annealed. Drilling is slow 
and therefore expensive; anneaHng is apt to change the plates 
and requires large expensive furnaces. Punching small and ream- 
ing is probably the best method. In this connection. Prof. A. 
B. W. Kennedy (see Proc. Inst. M. E., 1888, pp. 546-547) has 
called attention to the phenomenon of greater unit tensile strength 
of the plate along the perforations than of the original unperforated 
plate.* Stoney (" Strength and Proportion of Riveted Joints,'^ 
London, 1885) has compiled the following table: 

Table I. — Relative Percentage of Strength of Steel Plates Perforated 
IN Different Ways. 



Specimens. 


Unit Strength of Net Section between Holes compared 
with that of the SoUd Plate (loo Per Cent). 




i Inch. 


i Inch. 


i Inch. 


I Inch, 




Per Cent. 

lOI.O 

105.6 
113. 8 


Per Cent 

94.2 

105.6 

III. I 


Per Cent. 
82.5 

lOI.O 

106.4 


Per Cent. 

75-8 
100.3 
106. 1 


Punched and annealed 

Drilled 



For punched and reamed holes the same percentages may 
be used as for drilled. 

Professor Kennedy gives constants which may be obtained 
from, the following formula: 

Excess of unit strength of drilled steel plates in net section 
over unperforated section 

6.124^ 



-( 



2 + 



VJ 



4-5- 
2.5 



'-) 



per cent. 



/ is the thickness of plate in inches and r the ratio of pitch 

* This reported phenomenon is corroborated by tests made at Watertown 
Arsenal. See Tests of Metals, 1886, pp. 1264, 1557. It is fully explained by 
the condition of localized stress and the consequent prevention of lateral 
contraction. 



90 



MACHINE DESIGN. 



divided by diameter of hole. No data exist relative to iron 
plates in this matter. 

64. Kinds of Joints. — Riveted joints are of two general kinds : 
First, LAP-JOINTS, in which the sheets to be joined are lapped on 
each other and joined by a seam of rivets, as in Fig. 59a. Second, 
BUTT-JOINTS, in which the edges of the sheets abut against each 

Fig. 59. 



o 
o 





00 11 1 
00 1 1 



..M^^ ^ ^^ 





^^ 




Fig. 60. 



Other, and a strip called a "cover-plate" or "butt-strap" is riv- 
eted to the edge of each sheet, as in c. In recent years a lap-joint 



RIVETED JOINTS. 91 

with a single cover-plate has been used somewhat. It is shown 
in Fig. 59e. 

There are two chief kinds of riveting: Single, in which there 
is but one row of rivets, as in Fig. 59a; and double, where there 
are two rows. 

Double riveting is subdivided into "chain-riveting," Fig. 596, 
and "zigzag " or "staggered " riveting, Fig. ^gd. 

Lap-joints may be single, double chain, or double staggered 
riveted. 

Butt-joints may have a single strap as in Cj or double strap; 
i.e., an exactly similar one is placed on the other side of the joint. 
Butt-joints with either single or double strap may be single, 
double chain, or double staggered riveted. In butt-joints single 
cover-plates should be nine-eighths of the thickness of plates 
and double cover-plates five-eighths. 

To sum up, there are: 

f Single-riveted 

Lap-joints \ Double chain-riveted 

Double staggered-riveted 



Butt-joints. . . . 



f Single-riveted 

Single-strap -j Double chain-riveted 

[ Double staggered-riveted 
Single-riveted 

Double-strap -j Double chain-riveted 

Double staggered-riveted 



The demands of modem practice have added triple and quad- 
ruple joints to the foregoing. In high-pressure cylindrical boilers, 
for instance, common practice is to employ for the longitudinal 
seam the highly efficient joint shown in Fig. 60. Here we have 
a triple- riveted butt-joint with double cover-plates; on each side 
of the joint two rows of rivets are in double shear and one row, 
the outer, is in single shear. 

65. Failure of Joint. — ^A riveted joint may yield in any one of 
four ways: First, by the rivet shearing (Fig. 61 a); second, by the 
plate yielding to tension on the Hne AB (Fig. 61b); third, by the 



92 



MACHINE DESIGN. 



rivet tearing out through the margin, as in c; fourth, the rivet 
and sheet bear upon each other at D and E in d, and are both 
in compression. If the unit stress upon these surfaces becomes 
too great, the rivet is weakened to resist shearing, or the plate 
to resist tension, and failure may occur. This pressure of the 






Fig. 6i. 



rivet on the sheet is called "bearing pressure." It is obvious 
that the strongest or most efficient joint in any case will be one 
which is so proportioned that the tendency to fail will be equal 
in all of the ways. 

66. Strength of Materials Used. — ^As a preliminary to the 
designing of joints it is necessary to know the strength of the 
rivets to resist shear, of the plate to resist tension, and of the 
rivets and plates to resist bearing pressure. These values must 
not be taken from tables of the strength of the materials of which 
the plate and rivets are made, but must be derived from experi- 
ments upon actual riveted joints tested to rupture. The reason 
for this is that the conditions of stress are modified somewhat in 
the joint. For instance, in single-strap butt-joints, and in lap- 
joints, the line of stress being the center line of plates, and the 
plates joined being offset, flexure results and the plate is weaker 
to resist tension, the rivets in the mean time being subjected to 
tension as well as shear; if the joint yield to this stress in the 
slightest degree the "bearing pressure " is localized and becomes 
more destructive. The effect of friction between the surfaces 
of the plates under the pressure at which they are "gripped " 
by the rivets is another item of considerable importance. Ex- 



RIl^ETED JOINTS. 



93 



tensive and accurate experiments have been made upon actual 
joints and the results have been pubhshed.* 

The following table has been compiled as representing fair 
average results, and the values here given may be used for ordi- 
nary joints : 

Table II. — Values of ft, /«, and }c for Different Kinds of Joints. 



Kind of Joint. 



Lap-joint, single-riveted, punched 
holes 

Lap-joint, single-riveted, drilled holes. 

Lap-joint, double-riveted, punched 
holes 

Lap-joint, double-riveted, drilled holeb 

Butt-joints, single cover : Use values 
given for lap-joints. 

Butt-joints, double cover, single-riv- 
eted, punched holes 

Butt-joints, double cover, single-riv- 
eted, drilled holes 

Butt-joints, double cover, double-riv- 
eted, punched holes 

Butt-joints, double cover, double-riv- 
eted, drilled holes 

t Original plate 

t Original bar 



Iron. 



40000 
45000 

45000 
50000 



40000 

45000 

45000 

50000 
50000 



38000 
36000 

40000 
38000 



42500 
41000 
38000 
36000 

45000 



67000 
67000 

67000 
67000 



u 

89000 
89000 
89000 



Steel. 



60000 



47500 
45000 

48000 
46000 



fs' 

48000 

46000 
47500 
45000 
52000 



85000 
85000 

85000 
85000 



u 

I 00000 
I 00000 
I 00000 
I 00000 



The Master Steam Boiler-Makers' Assn., as the result of 
tests conducted by its committee, recommends, for iron rivets, 
/s = 42ooo and// = 4oooo; for steel rivets, /s= 46000, // = 44ooo. 

It will be noted that the values of jt are not given for steel. 
The tensile strength of steel varies through a considerable range 
due largely to differences in chemical constitution; it also follows 
a rough law of inverse proportion to the thickness of plates; 
i.e., thin plates will be almost sure to show higher tensile strength 



* See Proc. Inst, of Mech. Eng., 1881, 1882, 1885, 1888; Tests of Metals, 
Watertown Arsenal, 1885, 1886, 1887, 1891, 1895, 1896. Stoney's " Strength and 
Proportions of Riveted Joints," London, 1885. 

t If the original material varies from this, the values given above should be 
varied proportionatelv. 



94 MACHINE DESIGN. 

than thicker plates of the same composition. Furthermore, 
the method of perforation greatly affects the strength of the plates, 
as has been pointed out in § 63. Ordinary boiler-plates have a 
unit tensile strength ranging from 55,000 lbs. to 62,000 lbs. per 
square inch. For ordinary calculations ft may be taken as 
55,000 lbs. for punched plates and 60,000 for drilled plates. 
The shearing strength of rivets also varies inversely as their size, 
but these differences are slight. 

67. Strength, Proportions, and Efficiency of Joints. — No 
riveted joint can be as strong as the unperforated plate. The 
ratio of strength of joint to strength of unperforated plate is 
called the joint efficiency. 

As stated in § 65 the highest efficiency for a joint is obtained 
when the relations between thickness of plate, diameter of rivet, 
pitch, and margin are such that the tendency for the joint to 
fail in any one way does not exceed the tendency for it to fail 
in any other way. Formulae can be developed for finding their 
proper values for each form of joint. 

Let c^ = diameter of rivet-hole in inches; 
a = pitch of rivets in inches ; 
/ = thickness of plates in inches ; 

/« = tensile strength of plates in pounds per square inch; 
/^ = crushing strength of rivets or plates, if rivets are in 

single shear, pounds per square inch; 
//= crushing strength of rivets or plates, if rivets are in 

double shear, pounds per square inch ; 
/«= shearing strength of rivets in single shear, pounds per 

square inch; 
// = shearing strength of rivets in double shear, pounds 
per square inch. 

Each jcint may be treated as if made up of a successive series 
of similar strips, each unit strip having a width equal to a, the 
distance between centers of two consecutive rivets in the same 



RIVETED JOINTS. 



95 



row (see Fig. 62). If the stresses and proportions for one such 
strip are determined, the results obtained will, of course, apply 
to all of the others, and consequently to the whole joint. Con- 
sider such a strip of thickness / and width a. 




^^ Pitch V^ 




Fig. 62. 

Let P=ultimate tensile strength of unperforated strip, pounds; 
r= ultimate tensile strength of net section of strip, pounds; 
5= ultimate shearing resistance of all rivets in strip, 

pounds ; 
C= ultimate crushing resistance of all rivets or sides of 

holes, pounds; 
E = efficiency of joint. 
To illustrate this method, consider first the simplest joint, 
i.e., the single-riveted lap-joint. 

The unperforated strip has a tensile strength 

P=atjt (i) 

Along the row of rivets the net width of plate is less than the 
total width of the strip by an amount equal to the diameter of 
the rivet, and consequently the net tensile strength of the strip 
is expressed by the equation 

T = (a-d)tft . (2) 

In each unit strip there is but a single rivet with but one sur- 
face in shear, hence 

TT 



5 =-(f2/, = .7854^2/, 



(3) 



96 MACHINE DESIGN. 

The crushing resistance of the rivet, or of the plate around 
the hole, may be written as 

C = dtj, (4) 

For highest efficiency T=S^C. 

Equating S and C, (3) and (4), .'jS$4d^fs=dtfc. 

••• d = i.27tl^. . (5) 

This equation gives the proper theoretical value of d for a 
given value of /, and for materials represented by jc and /,. 
Equating T and S, (2) and (3), 

(a-d)tft = .-!&S4dV,- 

This gives the proper theoretical pitch. The efficiency of the 
joint is obtained by dividing T, 5, or C by P. 

In most cases the values of d and a as determined by (5) 
and (6) cannot be strictly adhered to. Stock sizes of rivets 
must be used in practice, and there are also limitations connected 
with the largest sizes it is convenient to drive. These equations, 
furthermore, do not take into consideration the stresses set up in 
the rivets when their shrinkage, due to cooling, is resisted by 
the plates, an item which may become excessive with the smaller 
diameters. The spacing of the rivets must also be modified quite 
frequently by the proportions of the parts to be connected, by 
allowance for proper space to form the heads, and by provision 
for tightness. In practice it is therefore often necessary to depart 
from these values. 

It must be borne in mind, however, that any departure from 
the values of d and a given in (5) and (6) destroys the equality 



RIVETED JOINTS. 97 

between T, 5, and C, and if such departure is made, the actual 
value of T, S, and C should be determined (by substitution of 
the values of d and a decided upon). The efficiency of the 
joint will then be found by dividing whichever has the smallest 
value, T, S, or C, by P. 

If the REAL efficiency of the joint is desired, the value of 
T must be obtained by increasing ft by the amount called for by 
the perforation of the plates. As explained in § 63 this will be 

( 2 +-^-F= ) l^ — j per cent greater than the ft of the original, un- 

perf orated plate. 

68. Problem. — The following problem illustrates the method 
of using Table II in connection with the formulas (5) and (6). 

What should be the dimensions of rivet-hole and pitch for a 
single -riveted lap-joint for |-inch iron plates using iron rivets? 

Table II gives as values of ft, fs, and fc 40,000, 38,000, and 
67,000 lbs. per square inch, respectively, for this form of joint, 

< = i"=-375"- 
Substituting these values, equation (5) becomes 



67000 ^ .. 
'i = i-27X.37SX^=.84" 



and equation (6) 



.7854 X. 8/ X 38000 

a = — -f .84 = 2.24 mches. 

.375X40000 

69. Proportions of Single-riveted Lap-joints. — Table III 
and Table IV have been computed in this way. As Table IV 
refers to steel joints, the values of ft, fs, and fc are 55,000, 47,500, 
and 85,000 lbs. per square inch, respectively. 



98 



Table III. 



MACHINE DESIGN. 

-Proportions of Single-riveted Lap-joints, Iron Plates, and 
Rivets, Punched Holes. 





fc 


.7854^2/5 


d 








- tn ^' 




TS 


.42 


1 . 12 






I 


•56 


1.49 






^ 


.70 


1.86 


f 


if 


1 


.84 


2.24 


f-f 


if -2 


i 


I. 12 


2.98 


f-M 


2-2A 


f 


1.40 


3-73 


f-i 


2-2f 


f 


1.68 


4-47 


i-ii 


2i-2f 


I 


1.96 


5.22 


i-i| 


2^-3 


I 


2.24 


5-96 


1-14 


2i-2| 


li 


2.52 


6. 71 


li-l 


2f-3 



Column I gives the thickness of plate; columns 2 and 3 give 
the corresponding calculated values of d and a for joint of maxi- 
mum efficiency; columns 4 and 5 give the values of d and a 
as compiled by Twiddell in the Proc. Inst, of M. E., 1881, pp. 
293-295, from boiler-makers' practice. It will be noted that the 
rivets used in practice (see column 4) are considerably smaller 
in diameter than those called for in column 2, and that this differ- 
ence grows more and more marked as the thickness of the plate 
increases. The reason for this is that the difficulty in driving 
rivets increases very rapidly with their size, i\ or i| inches 
being the largest rivet that can be driven conveniently. The 
equality of strength to resist bearing pressure and shear is there- 
fore sacrificed to convenience in manipulation. As the diameter 
of the rivet is increased the area to resist bearing pressure in- 
creases less rapidly than the area to resist shear (the thickness 
of the plate remainmg the same), the former varying as d and the 
latter as d"^; therefore if d is not increased as much as is neces- 
sary for equality of strength, the excess of strength will be to re- 
sist bearing pressure. If the other parts of the joint are made 
as strong as the rivet in shear, and this strength is calculated 
from the stress to be resisted, the joint will evidently be correctly 
proportioned. As machine-riveting comes into more general 



RIVETED lOlNTS 



99 



use and pneumatic tools are used in ** hand-work," this dis- 
crepancy will tend to disappear. 



Table IV. — Proportions of Single-riveted Lap-joints, Steel Plates, and 
Rivets, Punched Holes. 





^c 


.7854^2^5 , 






i 




"= «/, ^' 


d 


a 


^ 


•43 


1.08 


•47 


li 


\ 


•57 


1-45 


.61 


lA 


A 


•71 


1. 81 


.81 


2 


t 


.86 


2.17 


.94 


2A 


J 


1. 14 


2.89 


,..9 


3 



Column I gives the thickness of the plate; columns 2 and 3 
give the values of d and a calculated for joint of maximum effi- 
ciency; columns 4 and 5 give proportions from practice, the 
authority being Moberly (see Stoney, *' Strength and Proportions 
of Riveted Joints," p. 80). It will be noted how closely the 
theory and practice agree here for boiler joints.* 

70. Single-riveted Butt-joints. — To develop the general 
formulae for the values of a and d for single -riveted butt-joints 
with double cover-plates the same general method used in § 67 
applies. 

In this case the rivets are in double shear. Therefore 



5 = A', 
4 



while T = {a — d)tft, (2), as before and 

C = dt}/. 
Equating S and C, (7) and (8), 



(7) 



(8) 



* For further data, compiled from American practice, see sec. 73. 



lOO MACHINE DESIGN. 

and d = .64T}t (9) 

/« 

Equating T and S, (2) and (7), 

(a-d)tft = i.S7d2i/; 

--^-^V^ do) 

For Double-riveted Lap-joints the unit strip contains two rivets, 
each in single shear. The following equations cover the case : 

T=^(a-d)tft, 






C = 2dtfc 



d = i.2'jjt (11) 



71. Double-riveted Butt-joint. — For double riveted butt- 
joints, double cover-plate, either chain or staggered riveting, there 
are two rivets in double shear for each unit strip. 

T = (a-d)tft, 
4 



RIVETED JOINTS. lOi 

d = MT7t (13) 

Js 

3.14^^// , , , , 

^ = Jf^ +d (14) 

72. General Formulae. — The following general equations for 
riveted joints have been developed by Mr. W. N. Barnard:* 

The unit strip is of width equal to the pitch, the maximum 
pitch being taken unless all rows have the same pitch. 

The general expression for the net tensile strength of the unit 
strip is 

T = {a-d)tft (15) 

The general expression for resistance to shearing of the rivets 
in the unit strip is 

„ UTtd^ ^ 2mnd^ , , 

5=—/.+^-//, (16) 

in which n equals the number of rivets in single shear and m 
equals the number of rivets in double shear. 

The general expression for resistance to crushing of the unit 
strip is 

C = ndtj^ + mdtf/ (17) 

The tensile resistance of the solid strip is 

P = atft (18) 

Equating 5 and C, (16) and (17), and transposing, we get 

nf -\-fnf ' 

^ '-^■^^'i^/ ^-9) 

* See article, "General Formulas for Efficiency and Proportions for Riveted 
Joints," by Professor J. H. Barr in Sibley Journal of Engineering, Oct., 1900. 



I02 MACHINE DESIGN, 

Equating T and C, (15) and (17), 

,=(-/^')^+, (,,) 

Or, equating T and S, (15) and (16), 

The following equation for efficiency has been developed 
on the assumption that T = S = C. 

From £ = — we get, by substitution and transposition, 

E = . ...... (22) 

This equation is useful in finding the limiting efficiency ot 
joint for any form and materials; the actual proportions adopted 
may give a lower efficiency, but can never give a higher efficiency.'^ 
73. Proportions of Joints. — In American practice it will be 
found that there is more or less departure from the proportions 
which would be arrived at by the strict application of the prin- 
ciples laid down in the preceding articles. This variation is 
due to several considerations. Chief among them is the practical 
difficulty of driving large rivets, thus leading to the adoption of 
rivet diameters with reference to convenience of manipulation 
rather than efficiency of joint. As machines displace handwork 
the reason for this departure disappears and there is an increasing 
tendency to use the larger and more correct rivet diameters. 
Conservatism niust be reckoned with here and also in the failure 
to recognize the fact that rivet diameters do not depend solely 

* In the Proceedings of the Inst, of M. E., 1881, there is an article entitled 
"On Riveting, with Special Reference to Ship-work," M. Le Baron Clauzel, 
which enters deeply into the development of general formulas. 



RIVETED JOINTS. 103 

upon the thickness of plates, but also should vary with the kind 
of joint. Practice tends to hold to one diameter of rivet for each 
thickness of plate, irrespective of the kind of joint. 

Another item of practical importance is tightness against 
leakage under pressure. Most formulas are developed without 
consideration of this important factor. From a practical point 
of view, the joint fails when it begins to leak, actual rupture 
need not take place. The topic of the allowable maximum 
pitch as governed by experience with tightness of joints is 
discussed in § 80. 

The margin in a riveted joint is the distance from the edge of 
the sheet to the rivet hole. This must be made of such value that 
there shall be safety against failure by the rivet tearing out. 
There can be no satisfactory theoretical determination of this 
value; until recently it has been held that practice and experi- 
ments with actual joints showed that a joint would not yield in 
this way if the margin were made = d = diameter of the rivet 
hole. This is a safe rule for iron rivets in steel plates for any 
type of joint. Where steel rivets are used it will be well to 

increase this to — d. (See Power, Aug., 1905.) 
4 

The distance between the center lines of rows may be taken 

not less than 2.5^ for double-chain riveting, and 1.88^ for 

double-staggered riveting. This will insure safety against 

zigzag tearing of the plate, but brings the heads very close together. 

From these values and those of margins, as just discussed, the 

proper amount of lap can readily be determined for any kind of 

joint. 

74. Relative Efficiencies of Various Kinds of Joints. — The 

actual efficiencies of joints when tested show some departure 

from the calculated ideal efficiencies. The following Table (V) 

has been compiled from the results of tests to show roughly the 

relative efficiencies of various types of joints : 



104 



MACHINE DESIGN. 



Table V. — Relative Efficiency of Iron Joints. 



Efficiency 
Per Cent. 



Original solid plate 

Lap-joint, single-riveted, punched 

drilled 

double " 

Butt-joint, single cover, single-riveted. . 
" " " double-riveted, 

" double " single-riveted. 

" " " double-riveted 



45 
50 
60 

45-50 
60 

55 
66 



Relative Efficiency of Steel Joints. 



Efficiency Per Cent. 



Thickness of Plates. 

i-i i-i i-i 



Original solid plate 

Lap-joint, single-riveted, punched 

drilled 

* * double-riveted, punched 

drilled 

Butt-joint, double cover, single-riveted, drilled . . 
" " " double-riveted, punched. 

drilled... 



50 
55 
75 
80 
70 

75 
80 



45 
50 
70 

75 
65 
70 

75 



100 
40 
45 
65 
70 
60 

65 

70 



These tables are from Stoney's "Strength and Proportions of Riveted Joints." 

Triple and quadruple riveted butt-joints with double cover- 
plates show efficiencies ranging from 80 to 90 per cent.* 

75. Slippage. — ^At about 25 to 35 per cent of its ultimate load 
SLIPPAGE takes place in a riveted joint. This is probably due 
to the fact that at this load the friction between the plates, owing 
to the pressure exerted on them by the rivets, is overcome. It 
has been found the larger the cross-sectional area of the rivet 
the greater the percentage of ultimate load which can be with- 
stood without sHppage. It has also been found that large rivet- 
heads are better than small ones for the same reasom 



* For details of joints tested, see Tests of Metals, Watertown Arsenal, 1896- 



RIVETED JOINTS. 105 

The importance of the consideration of slippage has been fully 
established by the work of Professor Bach (" Die Maschinen- 
elemente," 9th ed. pp. 164-195). His careful and exhaustive 
experiments prove that : 

1. In cooHng the rivet shrinks away from the walls of the 
hole. 

2. In consequence of this, there is no tendency to shear off 
the rivet until after the joint has failed, for all practical purposes, 
by losing tightness because of slippage. 

3. The percentage of the ultimate or rupture load at which 
slippage takes place varies according to three items : 

a. It is directly proportional to the square of the diameter 

of the rivet. From this the desirability of using large 
rivets rather than small is further established. 

b. It is increased by calking, especially if both rivet heads 

are calked as well as the plate edges. 

c. It is greatly increased by holding the rivets under maxi- 

mum pressure until they are cool enough to have set. 

This gives better results than blows, light pressure, 

or early removal of pressure. 
Professor Bach argues that joints should not be proportioned 
with reference to the ultimate or rupture strength. He claims 
that the maximum pitch is determined by the condition of tight- 
ness against springing open between rivets when pressure is 
applied. The minimum pitch is that fixed by the spacing 
of rivet heads which is the least which will permit calking them. 
Between these limits he chooses pitch: 

1. So that the safe resistance to slippage (as experimentally 
determined by him) is equated to the stress on the joint due to 
the diameter of the vessel and the pressure. 

2. So that the unit stress in the plate between the rivets shall 
not exceed the safe working value of the plate material when the 
strength of the perforated section is equated to the stress due to 
the diameter and pressure. 



io6 



MACHINE DESIGN. 



Plate I shows graphically the proportions of riveted joints as 
determined from Professor Bach's formulas by M. Shibata in 
the American Machinist, Vols. 26 and 27. 

76. Rivet Size and Proportions. — In general the rivet should 
have a shank te inch smaller in diameter than the hole to be 
filled, while the head should have a diameter of from 1.6 to 2 
times the diameter of hole, and a height of from .6 to .75 time 
the hole diameter. Especial care should be taken in the case 
of machine-riveting to have just enough metal projecting beyond 
the hole to allow for the necessary upset for the shank to fill 
the hole, with just enough left over to fill the die for the head. 



Wl 



A 




l*^E-^ 






TABLE OF DIMENSIONS OF RIVET HEADS. 



Diameter 




Pan Head. 




Button Head. 


Counter Sunk. 


of Rivet. 




A 




B 


c 


d 


E 


F 


G 


E 


G 


E 


G 


1 


lA 


19 


^ 


I^ 


A 


II^ 


T^ 


ii 


IJ 


|i 


II 


'i 


^ 


lA 




f 


li 


If 


|i 


I4 


\E 


li 


f 


1 


lA 


T6 


f 


I]^ 


1 


if 


^ 


I 


If 


if 


11 


if 


' 


if 


h 



77. Problem. — How far must the tail of the rivet project in 
order to satisfy the above conditions for the following case: Two 
plates f inch thick each are to be connected, using |-inch rivets 
in xf-inch holes. The head is to be cone-shaped, having an out- 
side diameter of ij inches and a height of J inch. 

The cubical contents of the cone head = area of base X J 
altitude ==2.76 square inches X. 2 5 inch = .69 cubic inch. 



For Single. Double, Treble Riveted Lap Joint 
For SLngle Riveted Double Butt Strap Joint 
. " Double " " " " " 

" Treble " 




Thickness of Plates-T. 
Fig. 1. RIVET DIAMETERS BY BACH'S FORMULA. 





Fig. 3. SINGLE RIVETED LAP JOINT. 



Fig. 4. DOUBLE RIVETED LAP JOINT. ^ 




Fig. 7. DOUBLE RIVETED DOUBLE BUTT STRAP JOINT. Fig. 8. DOUBLE RIVETED DOUBLE 



PLATE I. 













































— ; 




n" 










































^ 


.\v^ 














-> 


D Y 


. 


















» 






y 


^, 


it" 












-4|k- 




















.kir' 




















j^,^ 




^ 


















^ 


k^ 




" 










,, 








i 


#--|;--- 


? 












/ 


^ 


^ 


<', 






















^ 


'^ 


i^H 


^ 










/: 


>: 


^ 


> 










— 




— 












"^ 


^J^ 








..^ 


:^ 


^ 


^ 


\^* 












« 6 

> 
















t 


^i3^e^ 


;^ 


b: 




















C! 














■.^ 


fe5?2 


^<' 






















O 3 












^p^ 


<X^^<^^^5fr^ 


























e. A" 










^ 


^y<^. 


^^^f^'"^ 


















1 






j' 








^C 


^ 


^^^ 


:^p 






























•v 




^ 


^ 


tj; 


^^ 


/"■i 
































-::>:%^ 


^^ 




































;%J^^ 


^n 




































^;;^ 


j-^ 


^ \ 


I 








































^ 












































^ 






1 










































1 




1 







































1" -Z'^ '6" 4" 5" 

Thickness of Plates-T. 

Fig. 2. RIVET LENGTH FOR GIVEN PLATE THICKNESS 





3.5. DOUBLE RIVETED LAP JOINT. 



Fig. 6. TREBLE RIVETED LAP JOINT. 




T STRAP JOINT. Fig. 9. DOUBLE RIVETED DOUBLE BUTT STRAP JOINT. 



RIVETED JOINTS. 107 

The difference in cubical contents between a hole tI inch in 
diameter by J inch long and a shank | inch in diameter and 
I inch long = fX. 7854(11^— 1^)=. 067 cubic inch. 

The amount required for head and upset therefore equals 
.69 + .067 = .757 cubic inch. 

The area of the finch shank = .60 square inch. .757 cubic inch, 

therefore, calls for a lensrth of -7^— =1.2=; inches. This amount 

^ .60 ^ 

would then be the projection through the plate. The length of 
rivet-shank called for would equal | inch + 1 J inches = 2 inches. 

Note. — Had the head been cup-shaped, its cubical contents 
should have been taken as that of a spherical segment. For cup- 
shaped heads the diameter is about 1.7 X diameter of hole, and 
the height about .6 X diameter of hole. The volume of the 
spherical segment is given by the following rule: Multiply half 
the height of the segment by the area of the base and the cube 
of the height by .5236 and add the two products. 

78. Countersunk Rivets. — Fig. 63^ shows the proportions for 
a countersunk rivet. C6untersunk rivets make a much weaker 
and less reliable joint than the ordinary form, and should only 
be used where it is absolutely necessary that the surface of the 
plate be free from projections. 

79. Nickel-steel Rivets. — Where peculiar conditions call for 
great strength of rivet combined with small area, it may be found 
desirable to use nickel-steel rivets. Experiments made by Mr. 
Maunsel White (see Journal Am. Soc. of Nav. Eng., 1898) on 
riveted joints using nickel-steel rivets showed an average shear- 
ing resistance of 85,720 lbs. per square inch for single shear, and 
an average of 90,075 lbs. per square inch for double shear. These 
values, it will be noted, are nearly double those of the very mild 
steel ordinarily used. The rivets were | inch in diameter, and 
some of the joints failed by tearing the plates, while others failed 
by shearing the rivets. 



ic8 MACHINE DESIGN. 

80. Construction of Tight Joints. — In general three types 
of riveted joints may be recognized : 

1. Those in which strength is the sole factor of importance, 
as in most purely structural iron and steel work. 

2. Those in which strength and tightness are equally deter- 
mining elements, as in boilers and pressure pipes. 

3. Those in which tightness is the prime consideration, as 
in tanks subjected to only light pressure. 

In punchi-^g holes in plates, it is, of course, necessary to 
have the hole in the die-block larger than the punch. The 
consequence is that the holes are considerably tapered and 
care should be exercised in joining the plates that the 
small ends of the holes be together as shown in Fig. 64^, 
and not apart as shown in Fig 6^B. It is obvious that at 
A the pressure on the rivet tends to draw the plates closer 
together, and that as the rivet cools its longitudinal shrinkage 
will tend to keep it a tight fit for the hole in spite of its diametral 
shrinkage. 




Fig. 64. Fig. 65. 

It is equally obvious that at B the pressure on the rivet will 
tend to force the plates apart and squeeze metal between them, 
and also that all shrinkage of the rivet will be away from the 
walls of the hole. 

In using drilled plates care must be exercised to remove the 
sharp burrs left by the drill, as experience has shown that this 
has a considerable effect on the strength of the joint. 

Where the plates form the walls of vessels to hold fluids, the 
joints must be designed with a view toward tightness as well as 



RIVETED JOINTS. 109 

Strength. For this purpose the edges are planed at a slight bevel, 
and calked as shown in Fig. 65 by a tool which resembles a cold- 
chisel with a round nose. Pneumatic tools are used for this pur- 
pose almost entirely, as they execute more uniform and rapid 
work than can be done by hand. In calking great care should 
be exercised not to groove the plates at A- A, as these are danger- 
points for bending, and an incipient groove is very apt to develop 
into a crack. It is largely on this account that the round- 
nose calking-tool has superseded the square-nose in the best 
practice. 

It has been found that the load which the joint will carry 
before leaking is greatly increased if both rivet heads are calked 
as well as the plate edges. (Bach's experiments.) 

The consideration of tightness has a determining effect on the 
maximum allowable pitch for any given thickness of plate and 
type of joint. Based upon practice the following values have 
been found safe for /^ '' plates: — 

Single riveted lap joints, pitch = yt 
Double riveted lap joints, pitch = 9.5t 
Double riveted butt joints, pitch (in outer row) = 14.5! 
Triple riveted butt joints, pitch (in outer row) = 2ot. 
Because of the use, with heavier plates, of rivet diameters 
which are proportionately too small, these ratios of pitch to 
thickness of plate will be found to decrease in practice as the 
thickness of plate increases. Thus for ^" plates they become 
5t, 6.6t, io.25t and i6t, respectively. 

81. Materials to be Used. — The material to be used in riveted 
joints depends, of course, on the nature of the work, but in gen- 
eral it may be said that extremely mild and highly ductile steel 
as free from phosphorus and sulphur as possible should be used. 
Open-hearth steel is greatly to be preferred to Bessemer.* 

* Standard specifications can be found in Kent's "Mechanical Eng. Pocket 
Book." The U S. Navy Dept. has adopted standard rules for riveting naval 
vessels. These can be found in Marine Engineering, Jan., 1898. 



^lo MACHINE DESIGN. 

82. Plates with Upsst Edges.— Some boiler-makers have 
adopted, as an expedient for saving material, a method of using 
plates with thickened (upset) edges. If we let / represent the 
thickness of the body of the plate and f the thickness of the edge, 
while a represents the pitch and d the diameter of hole, then, when 

a , 

t' = t -,, 

a — a 

the joint will be as strong as any other section of the plate, the 
joint being proportioned, of course, for the thickness t' . It is 
customary to thicken only the edges which form the longitudinal 
seam. This method is open to two serious objections. Unless 
the plates are very carefully annealed after being upset they are 
almost certain to be weakened by indeterminate working and 
cooling stresses. Moreover, although the original new joint may 
show as high an efficiency as if the plates throughout were of the 
thickness /', as corrosion proceeds, it acts more on the plate away 
from the joint than at the joint, because at the latter place the 
plate is protected by the cover-plate or rivet-head or both. The 
result is a shorter life under full pressure for the boiler with thin 
plates and thickened edges. 

83. Joints for More than Two Plates.— The joints considered 
thus far have dealt with the problem of connecting the edges 
of two plates only. In tanks and boilers which must have tight 
seams we are frequently confronted with the problem of joining 
three and even four plates. An instance is where the cross-seam 
and longitudinal seam of a boiler meet. The joint is made by 
thinning down one or more of the plates. Figs. 66 to 70 (taken 
from Unwin's "Machine Design") show the methods employed. 
Fig. 66 shows a junction of three plates, a, &, and c, where both 
seams are single-riveted lap-joints. It will be seen that the 
corner of a is simply drawn down to an edge and ** tucked 
under" c. 

Fig. 67 shows a junction of three plates where one seam is a 



RIVETED JOINTS. 



Ill 



single-riveted and the other a double-riveted iap-joint. As 
before, the comer of a is drawn down and tucked under c» 

Fig. 68 shows the junction of three plates where both seams 
are single -riveted, single cover- plate butt-joints. The plates 
merely abut against each other, but the longitudinal cover is 





oioo[ 


oi 

oi a\ 


r' 


-J" ■ 




Fig. 66. 



Fig. 67. 



Fig. 68. 



arawn down and tucked under the cross-seam cover which is 
thinned dow^n to match. 



a 

01 
\o\ 
01 
01 
ol 




^000 
D o o 

O O Q 




Fig. 69. Fig. 70. 

Fig. 69 also shows the junction of three plates; here the cross- 
seam is a single-riveted lap-joint while the longitudinal joint 
is a double-riveted butt-joint with double cover-plates. The 
upper cover-plate is planed on the end so that it can be tightly 
calked where it abuts against the plate c. 

A method of joining four plates is shown in Fig. 70. Both 
seams are single-riveted lap-joints, h and c are both drawn 
down as shown. 

84. Junction of Plates Not in Same Plane. — Where the plates 
to be joined are in different planes, it is customar}^ to use some 
one of the rolled structural forms. Fig. 71 shows the method 
of using an angle iron for plates at a right angle to each other. 



112 



MACHINE DESIGN. 



Where it is possible to turn a flange on one of the plates this 
method is often adopted. Care should be taken not to use too 



r: 



oooo 

^=i ^-"N /— -v 



Fig. 71. 





Fig. 72. 



sharp a radius of curvature (the inside radius must be greater 
than the thickness of the plate even with the mildest steel) and 
the flanged plate should be thoroughly annealed after it is bent. 

Fig. 72 shows the method of making flanged joints such as 
are frequently used in connecting boiler heads and shells. 

85. Problem. — The following problem will serve to illustrate 
the design of riveted joints for boilers. It is required to design a 
horizontal tubular boiler 48 inches in diameter to carry a work- 
ing pressure of 100 pounds per square inch. 

A boiler of this type consists of a cylindrical shell of wrought 
iron or steel plates made up in length of two or more courses or 
sections. Each course is made by rolhng a flat sheet into a 
hollow cylinder and joining its edges by means of a riveted joint, 
called the longitudinal joint or seam. The courses are joined to 
each other also by riveted joints, called circular joints or cross- 
seams. Circular heads of the same material have a flange turned 
all around their circumference, by means of which they are 
riveted to the shell. The proper thickness of plate may be 
determined from (I) The diameter of shell = 48 inches; (II) The 
working steam-pressure per square inch =100 pounds; (III) The 
tensile strength of the material used; let steel plates be used 
of 60,000 pounds specified tensile strength. 

Preliminary investigations of the conditions of stress in the 
cross-section of material cut by a plane (I) Through the axis; 
(II) At right angles to the axis, of a thin hollow cylinder, the 
stress being due to the excess of internal pressure per square inch. 



RiyETED JOINTS. 113 

Let L = the length of the cylindrical shell in inches; 
D = the diameter of the cylindrical shell in inches ; 
^ = the excess of internal over external pressure in pounds 

per square inch; 
/i=unit tensile stress in a longitudinal section of material 

of the shell due to p; 
J2 = unit tensile stress in a circular section of material of the 

shell due to p\ 
/ = thickness of plate; 
/i = ultimate tensile strength of plate. 
All stresses are in pounds per square inch. 

In a longitudinal section the total stress is equal to LDp, 

Dp 

and the area of metal sustaining it = 2Z/. Then /i= — - . 

TtD^p 

In a circular section the total stress = , and the area 

4 

sustaining it = 7rDt, nearly. Then 

^'~ 4 7iDt~ 4t' 

Therefore the stress in the first case is twice as great as in 
the second; and a thin hollow cylinder is twice as strong to 
resist rupture on a circular section as on a longitudinal one. 
The latter only, therefore, need be considered in determining the 
thickness of plate. Equating the stress due to ^ in a longitudinal 
section, and the strength of the cross-section of plate that sustains 

Dp 

it, we have LDp = 2Ltft. Therefore i=~~r, the thickness of plate 

that would just yield to the unit pressure p. To get safe thickness, 
a factor of safety K must be used. It is usually equal in boiler- 
shells to 5 or 6. Its value is small because the material is highly 
resilient and the changes of pressure are gradual, i.e., there are 
no shocks. This takes no account of the riveted joint, which 



114 MACHINE DESIGN. 

is the weakest longitudinal section, E times as strong as the 
solid plate, E being the joint efficiency =0.75 if the joint be 

double-riveted. The formula then becomes /= , ^ . Sub- 

2ttE 

stituting values, 

6X48X100 

/ = — — . — — =0.32 mch, say ^ mch. 

2X60000X0.75 

The circular joints will be single -riveted and joint efficiency 
will =0.50. But the stress is only one half as great as in the 
longitudinal joint, and therefore it is stronger in the proportion 
0.50X2 to 0.75, or I to 0.75. From this it is seen that a circular 
joint whose efficiency is 0.50 is as strong as the solid plate in a 
longitudinal section. From the value of t the joints may now be 
designed. 

Consider first the cross-seam. This is a single-riveted lap- 
joint. Assume drilled holes. 

Equations (5), 

Is 

and (6), .=^^^+<i, 

apply, while from Table II we get as values of ft, jay /c, 60,000, 

45,000, and 85,000 pounds per square inch respectively for steel 

plates and rivets. 

, Sqooo . , 

.*. d^i.2jX— X.3i25=.75mch 

' 45000 *^ ^ '-^ 

— 2 
.78t;4X.7S X 45000 o • ■, 

and a = — -^ -f .75 = 1.81 mches, say iH mches. 

.3125X60000 '^ > .7 lb 

The margm =^ = .75 inch. 

The lap of the cross-seam =3^ = 2.25 inches. 

The longitudinal seam will be a double staggered lap-joint. 



RIVETED JOINTS Hg 

Equations (ii) and (12) apply: 

J = i.27^/, and a = ^-^^P-' + d. 

From Table II, ft =60,000, /« =46,000, and /c =85,000; 

, 85000 . , 

/. (/ = i.27X^^^X.3i25=.74inch, say .75, 

— 2 
- 1-57 X. 75 X46000 . , . , 

and a = —7 + .75=2.93 inches, say 2^ inches. 

.3125X60000 '^ ^^ J / 16 

The distance between the rows = 1.88^ = 1.41 inches, say i^ 
inches. The total lap in the longitudinal joint =4.886^ = 3.66 
inches, say 3H inches. 

The joints are therefore completely determined, and a detail 
of each, giving dimensions, may be drawn for the use of the work- 
men who make the templets and lay out the sheets. 

Having determined the proportion of the joints, let these 
dimensions be used to calculate the actual efficiency of the longi- 
tudinal seam. 

Assume that the natural tensile strength of the unperforated 
plate is 60,000 lbs. per square inch. 

The excess of strength of drilled steel plates in net section 
over unperforated section (see § 63) is 



("if) 



jper cent. 



Here / = .3i25in. and r=^ =-^^ =3.92; 
/. excess due to perforation =3%, nearly. 



^^^ MACHINE DESIGN. 

60,000X1.03 =61,800 =/<. 
}a =46,000 from Table II. 

/,=85,ooo '' '' - 

r = (a-J)^/, = (2.9375-.75)(.3i25X6i,8oo) =42,250 lbs. 
S = 1.57^^/s = 1-57 X .75^ X 46,000 = 40,625 lbs. 
C = 2J//c = 2 X. 75 X. 3125X85,000 = 39,845 lbs. 
P = a/</ = 2.9375X6o,oooX.3i25 =55,075 lbs. 
Of T, 5, and C, the latter has the smallest value; the actual 

efficiency of the joint may be taken as =-5 = -^ = -7235 or 

72-35%- 

Since T, 5, and C are unequal it is evident that there has been 
departure from the conditions for maximum efficiency. There are 
two ways of restoring this equality, or at least diminishing the 
inequality. If a be slightly decreased, Tand P will be propor- 
tionately decreased and S and C will have the same values as 
before. 

Leaving a as before and increasing d, increases S as the 
square of d and C as d, while T and P remain as before. 

Inspection shows that T exceeds C by 5.7 per cent. There- 
fore a may be decreased by this percentage, or .17 inch. This 
is approximately ^ inch and reduces the pitch from 21I to 
2 1 inches. 

Using this value of a gives as the excess strength due to per- 
foration 4.33 per cent. 

.*. jt = 62,600 lbs., 

r = 39,i25 lbs., 
5 = 40,625 lbs., 
C = 39MS lbs., 
P = 5i,56o lbs., 

£=^ = ^-^ = 75.88%. 
P 51560 '^ 

The second method of balancing T, 5, and C would be by 



RIVETED JOINTS. Ii7 

increasing d. The next commercial size above | inch would 
be H inch. Leave a = 2\^ inches, and increase d io ^ inch, and 
first calculate the excess strength due to perforation. 

/ = .3i25 inch, r^-T--^^^. 
^ ^ ' d .8125 

Using these values, the excess = 4. 5 7 per cent. 

/< = 62,750 lbs., 

T = 4i,67o lbs., 

5 = 47.675 lbs., 

C = 43»i65 lbs., 

^ = 55.075 lbs., 
and, since T is least, 

£.^.41670^ 

P 55075 '^ 

The best result is that obtained by keeping d = \ inch, but 
changing a to 2f inches, and it would be advantageous to make 
these the proportions of the joints rather than those first de- 
termined. 

The next step is to check back, using the proportions de- 
cided upon, for the actual factor of safety which should not be 
less than 5. 

KDp 
From the equation (p. 107) t = ^-p, , we have 

2/t-tL 

. 2x6 2600x.7588x.3125 

.. A = 5— ^ = 0.10. 

48 X 100 

AttCxition should be called to the fact that the ordinary and 
less correct method of calculating efficiencies ignores the excess 



iiS MACHINE DESIGN. 

strength due to perforation, and the efficiency is simply taken 



as-p-. 



T_ 
p 

With a = 2if inches, and d = l inch, this would give us 



£ = ~^^ = 74.47%» instead of 72.35%. 



With a = 2| inches and d = l in^.h, it would give us 
£ = 1^ = 72.71% instead of 75.88%. 

With a = 2if inches and d=^ inch, it would give 
£ = |^ = 7i-97% 'nstead of 75.66%, 



CHAPTER VIII. 



BOLTS AND SCREWS. 



86. Classification and Definition. — Bolts and screws may be 
classified as follows: I. Bolts; II. Studs; III. Cap-screws, or 
Tap-bolts; IV. Set-screws; V. Machine screws; VI. Screws 
for power transmission. 

A ''bolt" consists of a head and round body on which a 
thread is cut, and upon which a nut is screwed. When a bolt 
is used to connect machine parts, a hole the size of the body of 
the bolt is drilled entirely through both parts, the bolt is put 
through, and the nut screwed down upon the washer. (See 
Fig. 73-) 




Fig. 73. 



Fig. 74. 



Fig. 75. 



A "stud" is a piece of round metal with a thread cut upon 

each end. One end is screwed into a tapped hole in some part 

of a machine, and the piece to be held against it, having a hole 

the size of the body of the stud, is put on and a nut is screwed 

upon the other end of the stud against the piece to be held. (See 

Fig. 74.) 

119 



I20 MACHINE DESIGN. 

A ''cap-screw" is a substitute for a stud, and consists of a 
head and body on which a thread is cut. (See Fig. 75.) The 
screw is passed through the removable part and screwed into a 
tapped hole in the part to which it is attached. A cap-screw 
is a stud with a head substituted for the nut. 

A hole should never be tapped into a cast-iron machine 
part when it can be avoided. Cast iron is not good material 
for the thread of a nut, since it is weak and brittle and tends to 
crumble. In very many cases, however, it is absolutely neces- 
sary to tap into cast iron. It is then better to use studs if the 
attached part needs to be removed often, because studs are put 
in once for all, and the cast-iron thread would be worn out 
eventually if cap-screws were used. 

The form of the United States standard screw-thread is 
shown in Fig. 76. The sides of the thread make an angle of 

60°. Instead of coming to a sharp 

H p — * .^^ 

A--r A--r /\^(k/\ point, the threads have a flat at top 
/ Iv/ \|/ \/ \ and bottom whose width is = J/?, p 
'"^^ being the pitch. Table VI gives the 

Fig. 76. standard proportions. 

For single threads the lead of the thread helix equals p, for double 
and triple threads it equals 2p and 3/?, respectively. If clockwise 
rotation of the screw causes the thread to enter the nut, the thread 
is termed right-hand; if counter-clockwise, left-hand. 

When one machine part surrounds another, as a pulley hub 
surrounds a shaft, relative motion of the two is often prevented 
by means of a "set-screw," which is a threaded body with 
a small square head (Fig. 77). The end is either rounded as 
in Fig. 77 a, or pointed as in Fig. 77 ^, or cupped as in Fig. 
77 c, and is forced against the inner part by screwing through 
a tapped hole in the outer part. 

Data relative to the holding power of set-screws will be 
found in § 109. 



BOLTS yIND SCREIVS. 



121 



Table VI. — U. S. Standard Screw-threads. 







Bolts and Threads. 




Hex. Nuts and Heads. 


iV and if. 


J, 


X. 


o 






"o 


i 


i 







^ 


u 

a; 


c 


c 


o 














p 






m 




^ 


rt 


S 


g 


% 

H 













"o 
o 








•^ . 






Q . 











'c) 


T) 


■g^ 




>> 


c g 


fcO 


-s 


to 


c 


c 


be 


£ 




^t 


:2 


c-g 


rti 




fS 


^t 







«§ 


n 


X. 


.r^ 


."2 


Sim 


SH 


%JA 


.cs 


%^ 


^ 


^ 


?c^ 


Q 


H 


Q 


< 


< 


C/3 


CO 


^ 


H 


H 


1 


Ins. 




Ins. 


Ins. 


Sq. Ins. 


Sq. Ins. 


Ins. 


Ins. 


Ins. 


Ins. 


Ins. 


Ins. 


i 


20 


■185 


.0062 


.049 


.027 


\ 


1^ 


11 


1 
4 


A 


tV 


A 


i8 


.240 


.0074 


.077 


•045 


If 


% 


tt 


A 


i 


if 


1 


i6 


.294 


.0078 


. no 


.068 




f 


¥ 


f 


A 




A 


14 


•344 


.0089 


.150 


•093 


M 


M 


TT7 


A 


1 


i?rV 


^ 


13 


.400 


.0096 


.196 


.126 


1 


tl 


I 


i 


1^ 


iH 


A 


12 


.454 


.0104 


.249 


.162 


31 


§f 


li 


A 


\ 


III 


f 


II 


•507 


.0113 


• 307 


.202 


13^ 


I 


I^ 


1 


^ 


li 


f 


lO 


.620 


.0125 


-442 


.302 


li 


ii^ 


II^ 


1 




i|f 


1 


9 


•731 


.0138 


.601 


.420 


lA 


If 


I§i 


\ 


if 


2^ 


I 


8 


•837 


.0156 


.785 


-550 


If 


lA 


4 


I 


if 


2if 


li 


7 


.940 


.0178 


•994 


.694 


lit 


If 


2^ 


li 


i^ 


2^ 


I 


7 


1.065 


.0178 


1.227 


•893 


2 


iM 


2^ 


li 


ii^ 


2f| 


ir 


6 


1. 160 


.0208 


1.485 


^•057 


21^ 


2i 


2^ 


If 


lA 


3^ 


li 


6 


1.284 


.0208 


1.767 


^•295 


2| 


2A 


2f 


A 




3fl 


if 


5^ 


1.389 


.0227 


2.074 


i^5i5 


2^ 


2i 


2M 


If 


lA 


3f 


If 


5 


1-491 


.0250 


2.405 


1.746 


2f 


2II 


31^ 


if 


lis 


3U 


i| 


5, 


1. 616 


.0250 


2.761 


2.051 


2if 


i 


3^ 


l| 


lif 


4^ 


2 


4^ 


I. 712 


.0277 


3-142 


2.302 


3i 


3A 


3f 


2 


lie 


4|| 


H 


4^ 


1.962 


.0277 


3-976 


3-023 


3i 


31^ 


4^ 


2i 


2^ 




2^ 


4 


2.176 


.0312 


4.909 


3-719 


zl 


3it 


4^ 


2^ 


2^ 


5fi 


2| 


4 


2.426 


.0312 


5 -940 


4.620 


A\ 


4^ 


4ff 


2f 


2ii 


6 


3 


3i 


2.629 


•0357 


7.069 


5.428 


4f 


4^ 


5f 


3 


2if 


6^ 


3i 


3^ 


2.879 


•0357 


8.296 


6.510 


5 


4it 


5^ 


3i 




71^ 


3^ 


3i 


3.100 


.0384 


9.621 


7-548 


5f 


5A 




2>\ 


3iC 


7ll 


3i 


3 


Z-Z^1 


.0413 


11.045 


8.641 


5f 


5H 


6fi 


3f 


3ii 


8i 


4 


3, 


3-567 


.0413 


12.566 


9-963 


6i 


6A 


7^ 


4 


3if 


8fi 


4i 


2| 


3-798 


•0435 


14.186 


11.329 


6| 


6t6 


7^ 


4i 


41^ 


9^ 


4i 


2f 


4.028 


•0454 


15.904 


12.753 


6| 


6if 


7!^ 


4i 


4A 


9f 


4f 


2f 


4.256 


.0476 


17.721 


14.226 


7i 


7i^ 


8M 


4f 


4i^ 


loi 


5 


2^ 


4.480 


.0500 


19-635 


15-763 


7f 


7^ 


8M 


5 


4if 


io|f 


5i 


H 


4-730 


.0500 


21.648 


17-572 


8 


7it 


9^ 


5i 


^^ 


iiff 


q^ 


2f 


4-953 


.0526 


23-758 


19.267 


8f 


8]^ 


9M 


5^ 


5i^ 


lis 


=^4 


2f 


5-203 


.0526 


25.967 


21.262 


8f 


8i| 


ic^ 


5f 


sii 


I2I 


6 


2i 


5-423 


■0555 


28.274 


23.098 


9i 


9A 


ic^ 


6 


5if 


I2i| 



122 



MACHINE DESIGN. 



The term ''machine screws" covers many forms of small 

screws, usually with screw-driver heads. All of the kinds given 

in this classification are made in great variety of size, form, 
length, etc. 





Fig. 77. Fig. 78. 

Thus far American manufacturers have failed to agree upoD 
standard dimensions for set-screws and machine screws,* 

For consideration of their design, we will divide bolts and 
screws into three classes : 

(a) Those which are put under no stress by screwing up. 

(h) Those which are put under an initial stress by tightening 

{c) Those which are used to transmit power. 

87. Analysis of Action of Screw. — Before taking these cases 
up in detail it will be well to examine into the general action 
of screw and nut. Reference is made to Fig. 79. The turning 
of a nut loaded with W lbs. may be considered as equivalent 
to moving a load W on an inclined plane whose angle with 
the horizontal is the same as the mean pitch angle of the 
thread a. 

* The report of the committee of the A. S. M. E. on this subject with suggeste^J 
standards, will be found in Vol. 28 of the Trans. A. S. M. E. 



BOLTS AND SCREIVS. 



123 



Let ri= outside radius of thread; 
^2 = inside radius of thread; 

r = mean radius of thread, approximately 

^ = pitch of thread; 

a = mean pitch angle, i.e., tan a = 



ri+r2 



2-Kr 

/£ = coefficient of friction between nut and thread; 
^ = angle of friction between nut and thread, i.e., tan ^ = /x. 

ist. To raise W . 

Consider PF as a free body moving uniformly up the incline 
under the system of forces shown in Fig. 80, where IT = axial 




Fig. 79. 



T 

Fig. 80. 



load, R the normal reaction between nut and thread due to W^ 
H the horizontal push forcing the nut up the incline, and F 
the friction in direction of incline due to the normal pressure R) 
then, by the ordinary laws of mechanics, 

R = W^co?>a, (i) 

F = pW ^cos a, (2) 

H = W idiii {a + cj>), ' (3) 

The turning moment M 

= Hr = Wri2in {a + 4>) (4) 



124 MACHINE DESIGN. 

P 
Since tan a = — , and tan <b = a. while 
27ir 

, . tan a + tan 6 

tan (a + 0) = — ; j, 

^ ^ I— tan a tan 

it follows that 

M=Hr=Wr^-^^ ; (5) 

2d. To lower W, 

H = W \.2.n {a-(f), (6) 

M=Hr = Wr^^^^^^-, ...... (7) 

The foregoing has applied to square threads. Consider V 
threads with /? = the angle of V with a plane normal to the axis 
of the screw. (See Fig. 81.) The mean helix angle =a as be- 




FlG. 81. 



fore, but now R slopes from W in two directions, making the 
angle a in the one, as before, but also making the angle /? with 
IF in a plane at right angles with the first. Hence 



R^W sec a sec /?. 
F = fiW sec a sec /: 



For raising load. 



Hr = Wr 



p + 27rrpi sec /? 
2nr — piistc /?* 



(8) 
(9) 

(10) 



DOLTS AND SCREIVS. 



12 



For lowering load, 



Hr = Wr 



p — 27zrpL sec /? 
2-r + piJ. sec /3* 



(II) 



88. Calculation for Screws Not Stressed in Screwing Up. — 

Returning now to {a). As illustrations of this class consider 
the eye-bolts shown in Fig. 82. It is customary to neglect the 
influence of the thread on the strength of the bolt, and to con- 
sider as the effective area, A, to resist stress, only the area of a 
circle w^hose diameter equals the diameter of the bolt at the 
bottom of the thread. In both cases considered a torsional 
stress is induced by screwing the engaging surfaces together, 
but if these surfaces are a proper fit, this stress is negligible, 
particularly since it exists only within the limits of the engaging 
threads where the action of the further working load which the 
bolt bears does not come into play. 




r^^r^ 



mm^m^. 



Fig. 82. 



u 

I 

T 

Fig. ^z- 



The eye-bolt being now subjected to the working load T in 
the direction of its axis, we have 



where / is the safe unit working stress for the material and con- 
ditions. 

If £/ = ultimate unit strength of the material, then, if the 



126 MACHINE DESIGN. 

load is a constant, dead load, / may be taken as great as — for 

o 
good wrought iron or mild steel. If the load is a variable one, 

slowly applied and removed, / should not exceed — for the 

same materials. If the load is variable and suddenly applied, 

U 

f should never exceed — for these materials, and in cases of 

shock may need to be much smaller than this. 

The case shown in Fig. 83 must not be confused with the 
preceding. Here (as will be explained under (b) ) there may be 
a tensile stress in A induced by compressing B-B between the 
shoulder and the nut. If the extension in the part A of the 
bolt due to the application of the force T later be greater than 
the compression caused in B-B by the tightening of the nut, 
then the shoulder will leave B-B, and simple tension = T results 
in all sections of the bolt below the nut, as in case (a). 

On the other hand, if the extension in the part A of the 
bolt due to the subsequent application of T be not so great as 
the original compression of B-B due to the screwing up, then 
we have in the part A a resultant tension greater than T. This 
case would come under (b). 

89. Calculation of Screws Stressed in Screwing Up. — (b) 
Combined tension and torsion are induced in a bolt by tighten- 
ing it. The stress may equal, or very greatly exceed, the tensile 
stress due to working forces. Consider the example shown in 
Fig. 78. Suppose the nut screwed down so that the parts con- 
nected by the bolt are held close together at E-F but have not 
yet been compressed. Suppose that the proportions are such 
that the wrench may be given another complete turn. The nut 
will move along the direction of axis of the bolt a distance = ^. 
The parts held between the head and nut will be compressed 
and the body of the bolt will be extended. 



BOLTS AND SCREIVS. 127 

The force applied at the point B, or end of the wrench 
(a distance / from the axis) will range from a value of o at the 
beginning of the turn to a value P at the finish. The average 

P 

value of the turning force will be approximately = —. 

This distance moved through by the point of application of 
the force is 2-I. Hence the work done in turning the nut a 
full turn under these conditions will be 



-.2rd = P7zl (12) 

2 



The resistances overcome by this application of energy are 
three in number: 

ist. The work done in extending the bolt. 

2d. The w^ork done in overcoming the frictional resistance 
between nut and thread. 

3d. The work done in overcoming the frictional resistance 
between nut and washer. 

These will be considered in order. 

ist. Let r = the final pure tensile stress in the bolt due to 

screwing up one turn, kt the beginning of the turn the ten- 

T 
sion = o. The average value may be considered =- — for the 

turn. The distance moved through by the point of application 
of this force in the direction of its line of action, in one turn = ^. 
The work done in extending the bolt 

T 
= -J • • • (13) 

2d. The frictional resistance between the threads of nut and 
bolt depends upon the form of the thread as well as the mate- 
rials used and the condition of the surfaces. (See equations (2) 



128 MACHINE DESIGN. 

and (9), § 87.) Assuming a V thread as being more commonly 
used for fastenings, the average value of the friction 

T 

F = fjr- sec a sec /? 

T 

(eq. (9) ), since the average load for the turn = — . 

The distance moved through by the point of application of 
F for one turn of the nut on the bolt = ^ cosec a. Hence the 
work done in overcoming the friction between bolt and nut in 

T 

the one turn = // — sec a sec /5 . p cosec a 



T 
= — }ip sec a sec /? . cosec a (14) 



3d. The frictional resistance between nut and washer due to 

T . T . 

a mean force — will be t/—, in which u! is the coefficient of 
2 ^2 

friction between nut and washer. The point of application of 

3 
this resistance may be taken at a distance of — ^i from the axis 

2 

of the bolt, ri being the outside radius of bolt-thread. The dis- 

tance moved through by the point of application for one turn of 

the nut = 27r|fi, and the work done in overcoming this frictional 

resistance 



=7/3^^1 • (15) 



Equating (12) to the sum of (13), (14), and (15), gives 



T T T 

P7tl = —p +—/jip sec a sec /? cosec a+— /i'37r/'ij 



BOLTS AND SCREIVS. 129 



whence 



^_ 2P7d 

P + p^p sec a sec P cosec a + fi^^Tzr I 

T . , 

ft=-j = nmi stress in bolt due to pure tension. . (17) 

In addition to this it must be borne in mind that the scre'w 
is subjected to a torsional moment whose value can be deter 
mined by considering the nut as a free body a> 
4^^^ \ shown in plan view in Fig. 84, where all of th( 
forces capable of producing moments about the 
axis of the nut are indicated as they exist at the 
end of the turn. 

Summing the moments about the axis of the 
bolt gives 
'^ Fig. 84. Hr = Pl-/T^r, (18) 

Hr is, of course, the torsional moment transmitted from the nut 
to the bolt. To find its numerical value substitute the value of 
T found in equation (16) and solve (18). 

The unit stress induced in the outer fibers of a rod of cir- 
cular section and radius ^2 ( = radius at bottom of thread) is 
found by means of the equation 

ts^ = M (19) 

J is the polar moment of inertia, in this case = ; c is the dis- 
tance from neutral axis to most strained fiber, in this case=r2; 
/s is the induced unit stress in outer fiber; M is the moment, in 
this case = .H'r. Combining equations (18) and (19) and sub- 
stituting these values gives 

/,= ^, . ...... (20) 



130 MACHINE DESIGN. 

The equivalent tensile unit stress due to the combined action 
of jt and js is found from the equation for combined tension 
and torsion, 



l = o.ssh+o.6sVh^ + 4ls^ (21) 

90. Problem. — What is the unit fiber stress induced in a 
U. S. standard J-inch bolt in screwing up the nut with a pull of 
one pound at the end of a wrench 8 inches long ? Arrangement 
of parts as shown in Fig. 78. 

In this case di = .500 in., ^1 = .25 in., 

^2 ="-400 in., r2 = .2 in., 
^ = .225 in., ^ = .126 sq. in., 

P 
a = angle whose tangent is — = 3° 7', 

sec a = 1. 0015, cosec a = 18.39, 

/? = 30°, sec ^=1.155, 

P = ilb., and / = 8 ins. 



T = 



From equation (16) 

2XiX^X 



.o77 + .o77Xo.i5Xi.ooi5Xi.i55Xi8.39 + o.i5X3X?rXo.25 
= 74.467 lbs. 
From equation (17), 

From equation (20), 

_ 2(1X8-0.15X74-467X1X0.25) 

Ja— 3 

;rXo.225 
= 213 lbs. 



BOLTS AhiD SCREirS. 131 

From equation (21), 

/=o.35X59i +0-65^591^ + 4X273^ 

= 691.4 lbs. 

If a pull of one pound on an 8-inch wrench applied to a J- 
inch bolt can induce a unit fiber stress of 691.4 lbs., since equa- 
tions (16) and (20) show that the stress increases directly as the 
pull, it follows that a pull of 30 lbs., such as is readily exerted 
by a workman, will induce a stress of 30X691.4 = 20,742 lbs. 
per square inch. 

91. Wrench Pull. — If this turning up be gradual and the 
bolt is not subjected to working stresses, this would be safe for 
either wrought iron or mild steel. On the other hand, if the 
final turning be done suddenly by means of a jerking motion or 
a blow, or a long wrench be used, or even an extra-strong grad- 
ual pull be exerted, there is evident danger of / having a value 
beyond the elastic limit of the material, even reaching the ulti- 
mate strength. 

It will be noticed also that the torsional action increases the 

fiber stress over that due to pure tension in the ratio of '—, i.e.. 

591 

in this problem, an increase of over 17 per cent. In general this 
increase will be from 15 to 20 per cent, depending chiefly upon 
the relation existing between p. and jj.\ It should also be noted 
that the pure tension, T, induced in the bolt by the moment PI 
may be taken as the measure of the pressure existing between 
the surfaces E-F (Fig. 78). In our problem this pressure, for 
P=3o lbs., would become 30X74.467 ^2234 lbs. 

As a general rule the length of wrench used by the workman 
is fifteen or sixteen times di, the diameter of bolt, and it may be 
stated that T = j^P for U. S. standard threads. 



MACHINE DESIGN. 



92. Calculation of Bolts Subject to Elongation. — Next con- 
sider the case shown in Fig. 85. Suppose that the nut is screwed 
up with a resulting tensile stress in the bolt 
= T. A working force Q tends to separate 
the bodies A and B at C-D. Assume that Q 
acts axially along the bolt. The question is, 
What value may Q have without opening the 
joiat C-D? 

A is the cross-sectional area of the bolt ; 

L is the original length between bolt-head 
and nut when A and B are just in 
contact at C-D but not compressed ; 

To is the tensile stress in bolt due to screwing up; 

?. is the total elongation of bolt due to To ; 

E is the coefhcient of elasticity of the bolt material. 

_, . unit strain i . . „ 

Then, smce ; _.^ ^^ ■ =-^, it follows that 




Fig. 85. 



unit stress £' 



j5_ 

A 



(i) 



In any given case this can be solved for X. 
Let A^ be the area of A and B compressed by the bolt action; 
X^ is the total compression (i.e., shortening) of A and B, 

due to the tightening of the bolt; 
Co is the total compressive stress which produces A'; 

.*. Co = To. 
B! is the coefficient of elasticity of the material A, B. Then 

)!_ 

Co ~E'' ' " 

A' 



(2) 



BOLTS ^ND SCREWS. 133 

This can be solved for X\ 

Now consider the condition when a working force, Q, acts 
tending to elongate the bolt so that A and B will just be ready 
to separate at C-D. In order that this separation may begin, 
the bolt, already elongated an amount X, must be elongated a 
further amount X\ 

For incipient separation the total elongation of the bolt then 
= }.+X\ and the total stress in the bolt corresponding to this 
elongation, = T', can be determined from the equation 



L I 



E (3) 



Considering the bolt-head as a free body (Fig. 86), it follows 

fTT] th2,t the forces acting on it at any instant will be, C, 

1 1 1 the reaction of the material of A due to its resistance 

^ to compression; Q, the working force; and T, the ten- 

FiG. 86. sion in the bolt. Hence 

T = C + Q (4) 

When the bolt is first screwed up, and Q=o, then C = T, 
and T = To, the tension due to screwing up. When Q comes 
into action, C is partly relieved, and when Q reaches such a 
value that the surfaces are about to part, then C=o and Q = 
T = T. (See equation (3).) 

An examination of these formulae shows certain facts which 
may be stated as follows: The tightness of the joint C-D de- 
pends upon the compressibility of A and B. 

Anything which increases the total compression, y, increases 
the tightness of the joint. This may be accomplished by in- 
creasing L or Co, or decreasing A\ It may also be increased 



134 



MACHINE DESIGN. 



by the introduction of a highly elastic body (i.e., gasket) between 
A and B. 

It also follows that the tension in the bolt when the joint is 
about to open, T\ must be greater than the tension due to 
screwing up, Tq, and therefore if Q be limited to a value equal 
to or less than To, there will be no opening of the joint. In 
general, A' is large compared with A, and k' very small com- 
pared with X, so that T' is not much greater than Tq. In order 
to be sure of a tight joint the initial tension should be taken 
To = 2(3. 

93. Problem I. — Calculate the bolts for a "blank" end for 
a 6-inch pipe using flanged couplings with ground joints, and 
no gaskets, as shown in Fig. 87. The excess internal pressure 
is to be 150 lbs. per square inch. 




The area subjected to pressure has a diameter of 7 J ms.; 



hence the total working pressure = 150 X 



^X7.5^ 



6627 lbs. 



The number of bolts is determined by the distance they may 
be spaced apart without danger of leakage due to the springing 
of the flange between the bolts. This distance may be taken 
equal to four or five times the thickness of the flange. In the 
problem under consideration, the diameter of the bolt circle will 
be approximately 9 ins., and using six bolts, the chord length 
between consecutive ones will be about 4J ins., which is per- 
fectly safe. 



With six bolts 



BOLTS AND SCREIVS. 135 



^ 6627 

e=-^^=iio5lbs. 



Take To = 2(2 = 2210 lbs. 

With a direct tension of 2210 lbs. due to screwing up, there 
is also the stress due to torsion. As stated in § 91, this may 
increase the fiber stress 20 per cent over that due to direct ten- 
sion. To allow for this the bolts used must be capable of safely 
sustaining a stress of 2210X1.20 = 2650 lbs. 

The allowable unit stress here may be taken rather high, 
since the conditions after once screwing up approximate a steady 
load. Assume steel bolts with an allowable unit stress of 
15,000 lbs. 

The area of each bolt at the bottom of the thread will then 

2650 

be =0.177 sq. in. This value lies between a i^-inch and 

15000 '' ^ 

a |-inch bolt. Select the latter with an area of 0.202 sq. in. To 
exert an initial tension of 2250 lbs. in a f-inch bolt would re- 
quire a pull of about 30 lbs. on a lo-inch wrench. (See § 91.) 
These values just about correspond to actual conditions in prac- 
tice. 

94. Problem II. — It is required to design the fastenings to 
hold on the steam-chest cover of a steam-engine. The opening 
to be covered is rectangular, io''Xi2''. The maximum steam- 
pressure is 100 lbs. per square inch. The joint must be held 
steam-tight. Studs of machinery steel having an ultimate ten- 
sile strength of 60,000 lbs. per square inch will be used. 

The total working pressure = 10 X 12 X 100 = 12,000 lbs. 

The number of studs to be used will be governed by the dis- 
tance they may be spaced without springing of the cover. The 
thickness of the latter being assumed to be | inch at the edge, 
the spacing should not exceed 5 Xf = -/'', say 4 inches. 



T36 



MACHINE DESIGN. 



-40^ 



The opening is 10'' X 12'', as shown in Fig. 8S. There must 
be a band about f or j inch wide 
around this for making the joint, Q p 

upon which the studs must not en- 
croach. This makes the distance 
between the centers of the vertical 
rows of studs about 14 inches, and 
between the horizontal rows about 
12 inches. Twelve studs can be used 
if arranged as shown in the figure. 

The greatest distance, that between rj q- 

the studs across the corners, will but Fig. 88. 

slightly exceed the allowable 4 inches. 



C 



C 



c 



c 



















With 12 studs, the working load on each = Q = 
To = 2Q = 2000 lbs. 



12000 
12 



1000 lbs. 



Allowing 20 per cent for torsional stress, increase this to 
2400 lbs. 

Allowing a unit stress of 15,000 lbs., as in Problem I, we 

have as the area of the stud at the bottom of thread = 0.160. 

15000 

This corresponds to a i^-inch stud. Since a workman may 
readily stress a bolt of this size beyond the elastic limit by exert- 
ing too great a pull in tightening, many designers would increase 
these studs to | inch or even J inch. 

95. Design of Bolts for Shock. — The elongation of a bolt 
with a given total stress depends upon the length and area 
of its least cross-section. Suppose, to illustrate, that the bolt. 
Fig. 89, has a reduced section over a length / as shown. This 
portion, A, has less cross-sectional area than the rest of the bolt, 
and when any tensile force is applied, the resulting unit stress 
will be greater in A than elsewhere. The unit strain, or elonga- 
tion, will be proportionately greater up to the elastic limit; and 



BOLTS AND SCREIVS. 



137 




if the elastic limit is exceeded in the portion A, the elongation 
there will be far greater than elsewhere. If there is much differ- 
ence of area and the bolt is tested to rup- 
ture, the elongation will be chiefly at A. 
There would be a certain elongation per 
INCH of A at rupture. Hence the greater 
the length of A, the greater the total elonga- 
tion of the bolt. If the bolt had not been 
reduced at A, the minimum section would 
be at the root of the screw-threads. The 
Fig. 89. Fig. 90. axial length of this section is very small. 
Hence the elongation at rupture would be small. Suppose there 
are two bolts, A with and B without the reduced section. They 
are alike in other respects. They are subjected to equal tensile 
shocks. Let the energy of the shock = £. This energy is di- 
vided into force and space factors by the resistance of the bolts. 
The space factor equals the elongation of the bolt. This is 
greater in A than in B, because of the yielding of the reduced 
section. But the product of force and space factors is the same 
in both bolts, '=£; hence the resulting stress in the minimum 
section is less for A than for B. The stress in A may be less 
than the breaking stress, while the greater stress in B may 
break it. The capacity of the bolt to resist shock is 

THEREFORE INCREASED BY LENGTHENING ITS MINIMUM SEC- 
TION TO INCREASE THE YIELDING AND REDUCE STRESS. This is 

not only true of bolts, but of all stress members in machines. 

The whole body of the bolt might have been reduced, as 
shown by the dotted lines in Fig. 89, with resulting increase of 
capacity to resist shock. Turning down a bolt, however, weak- 
ens it to resist torsion and flexure, because it takes off the material 
which is most effective in producing large polar and rectangular 
moments of inertia of cross-section. If the cross-sectional area 
is reduced by drilling a hole, as shown in Fig. go, the torsional 



138 MACHINE DESIGN. 

and transverse strength is but slightly decreased, but the elon- 
gation will be as great with the same area as if the area had 
been reduced by turning down. 

Professor Sweet had a set of bolts prepared for special test. 
The bolts were ij inches diameter and about 12 inches long. 
They were made of high-grade wrought iron, and were dupli- 
cates of the bolts used at the crank end of the connecting-rod 
of one of the standard sizes of the Straight-line Engine. Half 
of the bolts were left solid, while the other half were carefully 
drilled to give them uniform cross-sectional area throughout. 
The tests were made under the direction of Professor Carpenter 
at the Sibley College Laboratory. One pair of bolts was tested 
to rupture by tensile force gradually applied. The undrilled 
bolt broke in the thread with a total elongation of 0.25 inch. 
The drilled bolt broke between the thread and the bolt-head 
with a total elongation of 2.25 inches. If it be assumed that 
the mean force applied was the same in both cases, it follows 
that the total resilience of the drilled bolt was nine times as great 
as that of the solid one. "Drop tests," i.e., tests which brought 
tensile shock to bear upon the bolts, were made on other similar 
pairs of bolts, which tended to confirm the general conclusion. 

96. Problem. — It is required to design proper fastenings for 
holding on the cap of a connecting-rod like that shown in Fig. 
91. These fastenings are required to sus- 
tain shocks, and may be subjected to a 
maximum accidental stress of 20,000 lbs. 
There are two fastenings, and therefore 
each must be capable of sustaining safely 
a stress of 10,000 lbs. They should be 
designed to yield as much as is consistent 
with strength; in other words, they should 
be tensile springs to cushion shocks, and 
thereby reduce the resulting force they have to sustain. 




BOLTS AND SCRblVS. 139 

should therefore be used, and the weakest section should be 
made as long as possible. Wrought iron will be used whose 
tensile strength is 50,000 lbs. per square inch. The stress given 
is the maximum accidental stress, and is four times the working 
stress. It is not, therefore, necessary to give the bolts great 
excess of strength over that necessary to resist actual rupture 
by the accidental force. Let the factor of safety be 2. This 
will keep the maximum fiber stress within the elastic limit. 
Then the cross-sectional area of each bolt must be such that 
it will just sustain 10,000X2=20,000 lbs. This area is equal 
to 20,000^50,000=0.4 sq. in. This area corresponds to a 
diameter of 0.71 inch, and that is nearly the diameter of a 
J-inch bolt at the bottom of the thread ; hence f -inch bolts will 
be used. The cross-sectional area of the body of the bolt 
must now be made at least as small as that at the bottom of the 
thread. This may be accomplished by drilling. 

97. Jam-nuts. — When bolts are subjected to constant vibra- 
tion there is a tendency for the nuts to loosen. There are many 
ways to prevent this, but the most common one is by the use of 
jam-nuts. Two nuts are screwed on the bolt; the under one 
is set up against the surface of the part to be held in place, and 
then while this nut is held with a wrench the other nut is screwed 
up against it tightly. Suppose that the bolt has its axis vertical 
and that the nuts are screwed on the upper end. The nuts being 
screwed against each other, the upper one has its internal screw 
surfaces forced against the under screw surfaces of the bolt, and 
if there is any lost motion, as there almost always is, there will 
be no contact between the upper surfaces of the screw on the 
bolt and the threads of the nut. Just the reverse is true of the 
under nut; i.e., there is no contact between the under surfaces 
of the threads on the bolt and the threads on the nut. There- 
fore no pressure that comes from the under side of the under 
nut can be communicated to the bolt through the under nut 



I40 MACHINE DESIGN. 

directly, but it must be received by the upper nut and com- 
municated by it to the bolt, since it is the upper nut alone that 
has contact with the under surfaces of the thread. Therefore 
the jam-nut, which is usually made about half as thick as the 
other, should always be put on next to the surface of the piece 
to be held in place. 

98. Calculation of Screws for Transmission of Power. — 
(c) Screws are frequently used to transmit power. A screw- 
press is a common example, while the action of spiral gears, 
including worms and wheels, is that of screws and subject to 
the same analysis. 

The general action of screw and nut has already been treated. 
(See § 87.) 

With a square-thread screw it was found that the moment 

Pl=M, required to raise a load W, will he=Wr — (c) 

(page 124). 

This will induce a fiber stress /« = — j" (^^^ equation (19) ), 

W 

which must be combined with the tension, jt =-7-, in order to 

get the actual unit stress, /, remembering 



/ = .35^ + .65V/.' + 4/.^ (21) 

Let w=number of complete thread surfaces in contact 
between the nut and screw, and the projected area equals 

j[ 
n- (di^ — d2^) to bear the load W. 
4 

W=Kn-(d,^-d2^), 
4 

where K is the allowable pressure per square inch of projected 
thread area. 



BOLTS AND SCREIVS. 



141 



For nuts and bolts which are used as fastenings we may take: 
K = 2^oo lbs. for wrought or cast iron running on the same 

material or on bronze ; 
7^ = 3000 lbs. for steel on steel or bronze. 

With good lubrication, where the screw and nut are used to 

transmit power, we may take the values given in the following 

table : 

Table VII. 



Rubbing Speed in Feet 
per Minute. 



50 or less. 
100 

150 

250 

400 



Value of K. 



Iron. 



Steel. 



2500 


3000 


1250 


1500 


850 


1000 


400 


500 


200 


250 



The value of p. has been experimentally determined by Pro- 
fessor Kingsbury.* He concludes that for metallic screws turn- 
ing at extremely slow speeds, under any pressure up to 14,000 
lbs. per square inch of bearing surface, and freely lubricated 
before application of the pressure, the following coefficients of 
friction may be used. 

Table VIII. 



Lubricant. 


.". 




0. II 
0.143 

0.07 


Heavv-machinery oil (mineral) 


Heavy-machinery oil and graphite in equal 
volumes 





Regarding the efficiency of the square-screw thread to trans- 
mit energy, we may reason as follows : 

useful work 



The efficiency 



total work 



* Transactions A. S. M. E., Vol. XVII, pp. 96-116. 



142 MACHINE DESIGN. 

Wp 

which becomes for one turn = ^. 

2nrH 

wp 2nr W tan a tan a 

2nrH^ H ^T^ tan (a + 9^) ^tan (a: + ^)* * * ^^^^ 

From this it appears that e becomes o, for a = 0° and for 
a = 90° — (j), and must therefore have a maximum value between 
these hmits. To determine this maximum, write 



tan a , ,. 

tan a cot {a + 9; 



tan (a + ^j 
Taking the first differential and equating to o, 

cot (0: + ^) tan a 



cos^ a sin2 (^^ ^ ^) 
Solving which gives 



= 0. 



a=45°-^. 

^ 2 



To find the corresponding value of e, write (from (22) ) 
tan(45°-^) tan(45°-j) 



max. ^ 



tan(45°-7+'^) '^"(45° +7) 
To lower T^ with a square-threaded screw, 

Pl^Hr^Wr^-^^^^-!' (23) 

2-r + pfi ^ ^^ 

[The value of /.^ can be found from this, as explained in the 
section on raismg W, and combined with jt to obtain /.] 



BOLTS AND SCREIVS. 143 

Regarding the efiiciency in this case, 

if a< (f) the load will not sink of itself {i.e.^ overhaul), 

if a = </) we have a condition of equilibrium, 

if a >(/) the load will sink of itself {i.e., overhaul). 

For a screw which will not overhaul it becomes evident that 
the limiting value of a is <jS and the maximum efficiency 

tan a tan (j) i— tan^ 2 

e = - ; — — T-x =- 7 = =0.5—0.5 tan^ 29. 

tan {a + (j)) tan 2^ 2 j j 7 

The efficiency of a screw which will not overhaul can there- 
fore never exceed 0.5 or 50 per cent. 

For V threads, with ,5 = angle of V, with a plane normal to 
the axis of the screw for raising load, 

p^=tr /+^7^^^; , (,4) 

27ir — p[i sec /? ^ ^ 

which is evidently greater than (5), and 

tan a:(i — « tan a sec /?) , . 

6 = ^ ~~^ (2K) 

tan a + // sec /? > • • • • v d/ 

which is evidently less than e for square threads. (See equa- 
tion (22).) 

It is clear, then, that square threads should be used in pref- 
erence to V threads for screws for power transmission. 

For lowering the load with V threads 

27ir + pfi sec .5 ^ 

99. Problem. — Design a screw to raise 20,000 lbs. The 
screw must not overhaul. 

What moment need be exerted to lift the load ? 



144 MACHINE DESIGN. 

What will be the efficiency of the screw ? 
Select a square-thread screw of machinery steel running in 
a bronze nut. 

For a screw which w411 not overhaul a must be less than cj). 

.*. tan a< JUL. 

To be safe against overhauling with the materials used and 
good lubrication, // must not be given a greater value than o.io. 

.*. a<5°45' and = 5° 45'- 

The pure tension = 20,000 \hs. = W. In the preliminary calcu- 
lations, to allow for the effect of torsion, this will be increased 

. 25000 
so that / = — ^ — . 

In this equation / is the allowable unit stress in pounds per 
square inch, and A is the area of the screw at the bottom of the 
thread in square inches. 

Assume that this screw is frequently loaded and unloaded, 

and not subject to shocks nor reversal of stress so that / = 

25000 

12,000 lbs. per square inch for mild steel. Then A = — = 

^ ^ 12000 

2.08 sq. ins. This corresponds to a diameter of if inches at 

bottom of thread. 

P 
From tan a = , we have 

27ir 

p = 27ir tan a. 
Remembering that for square threads the depth of the thread 
= — , and that ^2 is the radius at the bottom of thread, and 



2 



'. r = r2 + —,it follows that 
4 



BOLTS AND SCREWS, I4S 

/ P\ 

p=27z[r2 + —\ tan a, 



27Z tan a 
p p = 27tr2 tan a, 



27[r2 tan a 

p=-^ — 

I—— tan a 

2 



/. ^ 



;i25Xo.i ... , 

-— = .606 inch. 



1.57X0.1 



This is not a thread to be easily cut in the lathe. It would 
be desirable to modify the value of p so that the thread can be 
readily cut. It is obvious that p cannot be increased without 
increasing r proportionately, else a will have a greater value 
than is allowable. It will be more economical to reduce p. 
The nearest even value would be J inch, and this will be selected. 
Check this for strength: 
c?2"= 1-625 inches, ;'2 =0.8125 inch, _/> = 0.5 inch. 

c?i = 2.125 inches, ri = 1.0625 inches, 
^ = 1.875 inches, /- =0.9375 inch, 

tan q:= — =— — -^f^ = 0.087, 

27:r 2X71X0.9375 

which is safe, as it is less than the value of // = o.io. 
From equation (5), the moment, 

27zr — ppL 

. „, ,, 0.5 + 2X7CX0.9375XO.IO 

.*. P/ = 2ooooXo.93757— ;'— r^^ 

^^^^(2X7rXo.9375) -(0.5x0.10) 

= 3496 in. -lbs. 



146 MACHINE DESIGN, 

From equation (19), the fiber stress due to torsion, 



cPl 2PI 
J ~7zy^ 



h J ^,, s- 



. r 2X3496 „ . , 

• • h=:^Q ii^^^^^^^ P^^ square mcho 
From equation (17), the unit stress due to tension, 

, W 20000 -^ . , 

A=]4-^^:^^95^'' lbs. per square mche 

From equation (21), the combined stress, 



/ = o.35/, + o.65V/,2 + 4/,2. 



•'• / = o-35X9597+o.65V95972 + 4X4i4o2 
= 12,379 lbs. per square inch, 

which is near enough 12,000 to be considered safe. 
The efficiency, from equation (22), 



tan a 
e = 



tan (0; + ^)* 

Since tan a = 0.087, ^ ^ 5°- 
Since ji = tan 9S = o. 10, ^ = 5° 45'. 

/., q: + (^ = io°45', tan 10° 45^ = 0.1899. 
•■•' = 5:^9 = °-4S8i'°'^ 45-81%. 
The height of the nut is determined from the equation 



4 



BOLTS AND SCREIVS. 147 

in which W is the load, n the number of complete threads in 
the nut, di the outside and 62 the inside diameter of thread. 
K is the allowable pressure in pounds per square inch, and its 
value depends upon the speed. See table in sec. 98. 

Assuming the screw to have a rubbing velocity of less than 
50 feet per minute, K = 3000. Then 

W 20000 

n = - 



^oo^{d,^-di) 3000X0.7854(2.125^-1.6252) 

= 4.5, nearly. 
The height of nut = ^Xw = J"X4.5" = 2i". 



CHAPTER IX. 

MEANS FOR PREVENTING RELATIVE ROTATION. 

100. Classification of Keys. — Keys are chiefly used to pre- 
vent relative rotation between shafts and the pulleys, gears, etc., 
which they support. Keys may be divided into parallel keys, 
taper keys, disk keys, and feathers or splines. 

loi. Parallel Keys. — For a parallel key the "seat," both 
in the shaft and the attached part, has parallel sides, and the 
key simply prevents relative rotary motion. Motion parallel to 
the axis of the shaft must be prevented by some other means, 
as by set-screws which bear upon the top surface of the key, as 
shown in Fig. 92. A parallel key should fit accurately on the 
sides and loosely at the top and bottom. The following table 
(IX) for dimensions for parallel keys is from Richards 's "Manual 
of Machine Construction." 

Table IX. 

Diameter of shaft = (/= I ij i^ if 2 2^ 3 3i 4 

Width of key =w= ^ ^^iiMM^^ii 
Height ofkey=;=T^ i A I A I ^ f I 

Excellent parallel keys are made from cold-rolled steel with- 
out need of any machining. 

102. Taper Keys. — A taper key has parallel sides and has 
its top and bottom surfaces tapered, and is made to fit on all 
four surfaces, being driven tightly "home." It prevents rela- 
tive motion of any kind between the parts connected. If a key 
of this kind has a head, as shown in Fig. 93, it is called a "draw 

key," because it is drawn out when necessary by driving a 

148 



MEANS FOR PREVENTING RELATIVE ROTATION. 



149 



wedge between the hub of the attached part and the head of the 
key. When a taper key has no head it is removed by driving 
against the point with a *' key-drift/' 




Fig. 92. 



Fig. 93. 



Fig. 94. 



John Richards's rule for keys is (see Fig. 94) w 



t has 



such value that a =30°. This rule is deviated from somewhat, 
as shown by the following table (X) taken from Richards's 
"Manual of Machine Construction," page 58: 



li 






li 2 
■h 



Table X. 

2i 3 3i 

I f f I 



5678 
li if i| if 



When two or more keys are used, 'w=d^6, t being, as before, 
of such value that a shall =30°. 

The taper of keys varies from \ to \ inch to the foot. 

103. Fitting Shaft and Hub. — In using taper keys it is cus- 
tomary to bore out the hub slightly larger than the diameter of 
the shaft so that the wheel may be readily removed after the 
key is withdrawn. This allowance in diameter should not be 
greater than that for a running fit, say, 

1000 ' 



in which formula A is the difference in diameter between the 
bore of hub and size of shaft, expressed in decimal parts of an 



it;o 



MACHINE DESIGN. 



inch, and D is the nominal diameter of shaft in inches. Where 
the parts do not have to be taken apart frequently, it is vastly 
better to use a driving fit, i.e,, to bore out the hub smaller than 

D 



— + 0.5 
1000 



and to use 



the diameter of the shaft by an amount A 

parallel keys. 

Where a single taper key is used the effect is to make the 
wheel and shaft eccentric, as can be seen in Fig. 95. The bear- 
ing is limited to two points. A, B, and the connection is unstable 
for the transmission of power. 





Fig. 95. 



Fig. 96. 



If great care is not exercised in having the taper of keyway 
exactly the same as the taper of the keys, a further difficulty 
arises in that the wheel will be canted out of a true normal plane 
to the shaft-axis. This can be seen in Fig. 96. 

By using two keys, placed a quarter or third of the circum- 
ference apart, a much more stable connection is obtained, as it 
will give three points of bearing. A, B, and C. <See Fig. 97.) 
Eccentricity is not avoided by this method. 

Another method is that shown in Fig. 98. Here the hub is 
bored out somewhat larger than the shaft and a bushing B is 
used opposite the key. This bushing has its outside diameter 
exactly the diameter of the bore, and its inner diameter exactly 



MEANS FOR PREi^ENTING RELATIVE ROTATION. 



151 



the diameter of the shaft, and it surrounds slightly less than 
half the shaft. When the key is driven home this insures the 





Fig. 97. 



Fig. 98. 



wheel and shaft being concentric, but extra care must be taken 
that key and keyway have precisely the same taper. Three 
keys uniformly spaced around the shaft also give a means of 
making the wheel and shaft concentric. This method is rarely 
used. 

104. Woodruff Keys. — The Woodruff or disk system of keys 
is used by some manufacturers. The key is a half disk, as can 
be seen in Fig. 99. Under this system the keyway is cut lon- 
gitudinally in the shaft by means of a milling-cutter. This cut- 
ter corresponds in thickness to the key to be inserted, and is of 





Fig. 99. 



Fig, 100. 



a diameter corresponding to the length of the key. The key 
being semicircular, it is sunk into the shaft as far as will allow 
sufficient projection of the key above the surface to engage the 
keyway in the hub. 



152 



My4CHINE DESIGN, 



Owing to its peculiar shape the key may be slightly inclined, 
so that it will serve to support the wheel on a vertical shaft, pro- 
vided the key-seat in the hub is made tapering and of the proper 
depth. 

104. Saddle, Flat and Angle Keys. — Saddle keys (Fig. 100, A) 
and keys on flats (Fig. 100, B) are used occasionally. They 
have not the holding power of sunk keys. 




Fig. 100. 

The type of key shown in Fig. 100 C has much to be said in 
its favor both as regards ease and accuracy in obtaining a stable 
connection and also as regards suitability of form to resist stresses. 
It will be noted that the surfaces are normal to the lines of action 
of the forces transmitted. The pressure per square inch should 
not exceed 17,000 pounds. The height of key is taken equal 
to 0.2 diameter of shaft. 

The Kernoul key, shown in Fig. 100 D, is for use in driving 
in only one direction. A portion of the hub is cut out so as to 



MEANS FOR PREVENTING RELATIVE RELATION, 153 

form an eccentric slot. In this the key fits as shown. The 
inner face of the key, curved to the radius of the shaft, should 
be left rough so as to seize the shaft, while the outer face, curved 
to fit the slot in the hub, is smooth finished. When the shaft 
rotates (in this case) counter-clockwise, the resistance to the 
hub's motion being then as indicated by the arrow, the surface 
of the slot tends to slide up on the key, causing it to wedge in 
between the shaft and hub, forming a firm connection. When 
the shaft rotates in the opposite direction and the resistance 
to the hub's motion is reversed, the slot of the hub tends to leave 
the key, relieving the pressure and permitting easy removal from 
the shaft. At " a " and "b " there are counter-sunk screws for 
setting up and loosening the key. In Fig. 100 E is shown a special 
form of this type of key. It is known as the Barbour key and 
is chiefly used for fastening the cams on the shafts of stamp 
mills.* 

In the study of keys which drive in one direction only it is 
proper to include the roller ratchet. The simplest form is shown 
in Fig. 100 F. The hub is recessed as shown and the roller R 
placed in the recess, held in position by a spring. The direction 
of shaft rotation and hub resistance being as shown the roller 
becomes wedged between the two, forming a driving connection. 
With reversal of direction the roller is freed and the shaft and 
hub may have relative motion. 

Generally more than one roller is used and the mechanism 
takes the form shown in Fig. 100 G. ^ is a hardened and 
ground steel ring or bushing and B should also be hardened. 
Each roller should be held in place by a spring as shown at C, 
Such ratchets permit of rapid reciprocation. Complete details 
and descriptions of further clutches of this type will be found in 
the American Machinist of Dec. 21, 1905. 

* Patent held for this purpose by the Risdon Iron Works. The distinguishing 
feature from the plain Kernoul key lies in the use of the inside projecting tongue 
which fits in a ke>'^^•ay cut in the shaft. 



154 MACHINE DESIGN. 

105. Strength of Sunk Keys. — The strength of the latter is 
the measure of their holding power. A key of width =w, thick- 
ness =/, length =/, unit shearing strength =/«, and unit crushing 
resistance = jc will have a shearing strength = /^w/ and a crush- 
ing resistance /cj//. 

If r= radius of the shaft, the moment which the key can 
resist will be measured by rwljs or Jr///^, whichever is smaller 
in value. 

All dimensions being expressed in inches and resistance 
in pounds per square inch, the moments will, of course, be 
expressed in inch-pounds. Experiments made by Professor 
Lanza indicate that the ultimate value for /^ for cast iron = 
30,000 lbs., for wrought iron = 40,000 lbs., and for machinery 
steel = 60,000 lbs. A factor of safety of 2 would be advis- 
able with these values. 

106. Feathers or Splines are keys that prevent relative rota- 
tion, but purposely allow axial motion. They are sometimes 



Fig. 10 1. Fig. 102. 

made fast in the shaft, as in Fig. loi, and there is a key *'way" 
in the attached part that slides along the shaft. Sometimes the 
feather is fastened in the hub of the attached part, as shown 
in Fig. 102, and slides in a long keyway in the shaft. 

It is frequently undesirable to have the feather loose. In 
such cases it is common to use tit-keys as shown in Fig. 103. 
The keys may be fastened to either hub or shaft. The tits 
are forged on the keys. Corresponding holes are drilled 
and countersunk in the piece to which the key is to be fastened. 



MEANS FOR PREVENTING RELATIVE ROTATION. 



55 



and after the key is placed in position the ends of the tits are 
riveted over to hold it securely in place. 

Machine screws are sometimes used in place of tits, but 
they suffer from the disadvantage of jarring loose. 

A satisfactory way of holding a key in a hub is shown in 
Fig. 104. 

Where the end of a stud is to receive change-gears a con- 
venient form of key is the dovetail shown in Fig. 105 in cross- 




FiG. 103. 



Fig. 104. 



Fig. 105. 



section. The dovetailed key-seat is generally cut with a mill- 
ing-cutter, and is made a tight fit for the key. After the latter 
is in place the shaft is calked against it at A- A. 
For feathers, Richards gives: 



w = \ 







Table XI. 










li 




2I 2\ 


1 


1 

i 


4 


4i 
1 

8 



107. Round Taper Keys. — For keying hand-wheels and 
other parts that are not subjected to very great stress, a cheap 
and satisfactory method is to use a round taper key driven 
into a hole drilled in the joint, as in Fig. 106. If the two parts 
are of different material, one much harder than the other, 
this method should not be used, as it is almost impossible in 
such case to make the drill follow the joint. For these keys 



156 



MACHINE DESIGN. 



it is customary to use Morse standard tapers, as reamers are 
then readily obtainable. 

io8. A Cotter is a key that is used to attach parts that are 
subjected to a force of tension or compression tending to sepa- 
rate them. Thus piston-rods are often connected to both pis- 
ton and cross-head in this way. Also the sections of long 
pump-rods, etc. 

Fig. 107 shows machine parts held against tension by cot- 





FlG. 106. 

ters. It is seen that the joint may yield by shearing the cot- 
ter at AB and CD, or by shearing CPQ and ARS; by shear- 
ing on the surfaces MO and LN; or by tensile rupture of the 
rod on a horizontal section at LM. AD of these sections should 
be sufficiently large to resist the maximum stress safely. The 
difficulty is usually to get LM strong enough m tension; but 
this may usually be accomplished by making the rod larger 
or the cotter thinner and wider. It is found that taper sur- 
faces if they be smooth and somewhat oily will just cease to 
stick together when the taper equals 1.5 inches per foot. The 
taper of the rod in Fig. 107 should be about this value in order 
that it may be removed conveniently when necessary 

From consideration of the laws of friction it is obvious that 
where a taper cotter is used, either alone as in Fig. 108 or in 
connection with a gib as in Fig. 109, the angle of taper a must 
not exceed the friction angle (f>. That is, if the coefficient of 
friction be /i, then pt = tan and tan a must be less than tan (f> 



MEANS hOR PREVENTING RELATIl^E ROTATION. 



157 



or ft. Since, for oily metallic surfaces, n may have a value 
as low as 0.08, it follows that a must not exceed 4^°. If both 
surfaces of the cotter slope with reference to the line of action 
of the force, the total angle of the sloping sides must not 
exceed 9°. 



z=^ 



U-- 




Fig. 108. 



Fig. 109. 



109. Set-screws (see § 86, p. 120) are frequently used to pre- 
vent relative rotation. They are inadvisable for heavy duty. 
Experiments made by Professor Lanza* with f-inch wrought- 
iron set-screws, ten threads to the inch and tightened with a 
pull of 75 lbs. at the end of a lo-inch wrench, gave the 
following results : 

Table XII. 

v..^A ^t T?^A Holding Power at 

Kind of End. Surface of Shaft. 

Ends perfectly flat, ^ inch diameter, . . Average 2064 lbs. 

Rounded ends, radius i inch • " 2912 " 

Rounded ends, radius ^ inch " 2573 " 

Cup-shaped and case hardened " 2470 " 

1 10. Shrink and Force Fits. — Relative rotation between 
machine parts is also prevented sometimes by means of shrink 
and force fits. In the former the shaft is made larger than 
the hole in the part to be held upon it, and the metal surround- 
ing the hole is heated, usually to low redness, and because of 
the expansion it may be put on the shaft, and on cooling it 

* Trans. A. S. M. K., Vol. X. 



158 MACHINE DESIGN. 

shrinks and '' grips " the shaft. A key is sometimes used in 
addition to this. For proper allowances see the accompanying 
table. The coefficient of linear expansion for each degree 
Fahrenheit is 0.0000065 for wrought iron and steel and 
0.0000062 for cast iron. Low redness corresponds to about 
600° F. and therefore causes an expansion of the bore of about 
0.004 inch per inch of diameter. 

Force fits are made in the same way except that they are 
put together cold, either by driving together with a heavy sledge 
or by forcing together by hydraulic pressure. The necessary 
allowance, i.e., excess of shaft diameter over the diameter of 
the hole, is given in the following table (XIII), compiled by 
Mr. S. H. Moore.* 

Column I gives values of ''pressure factors" which are to 
be used in connection with forced fits to determine the pressure 
necessary to force the machinery steel shaft into the cast-iron 
hub. The formula to be used is 

Area of difference in dia. be- p,^ 

^ . surface of fit tween plug and bore 

Pressure m tons = ■• 

2 

In using this table for forced fits the surfaces should be as 
smooth as possible; if they are ground it is best. They should 
be well lubricated. Mr. Moore recommends linseed-oil for this 
purpose. 

Compared with average practice the forcing-fit allowances of 
this table are too large. Satisfactory results will be obtained 
by using just one half these tabular values. 

In general, pressure fits are not employed on diameters ex- 
ceeding 10 ins., shrinkage fits being used for large work. 

* Trans. A. S. M. E., Vol. XXIV. 



MEANS FOR PREVENTING RELATIVE ROTATION 159 

Table XIII. — Data Relative to Fits and Fitting. 



Pressure 


Nominal Yo\ 


1 
1 

-cing Fit, Sh 


rinking Fit, 


Driving Fit, Rur 


ining Fit, 


Factors, 
PF. 


"Tt^ ^" 


owances. i^ 


lUovvances. 


Allowances. All 


owances. 




1 






.0006 


00058 
ooo6n 




4 






.ODOS 




1 






.0009 


00073 






391 


I 


.0025 


0016 


.0010 


0008 


319 


i| 


•0035 


0021 


.0013 


0010 


240 


2 


.0045 


0026 


.0015 


OOII 


156 


3 


.0065 


0037 


.0020 


0014 


"5 


4 


.0085 


0048 


.0025 


0018 


91 


5 


.0105 


0058 


.0030 


0021 


l^ 


6 


.0125 


0069 


•0035 


0024 


64 


7 


-0145 


0079 


.0040 


0027 


^l 


8 


.0165 


0090 


.0045 


0030 


48.5 


9 


•0185 


OIOI 


0050 


0033 


43 


10 


.0205 


OIII 


•0055 


0036 


39 


II 


0225 


0122 


.0060 


0039 


36 


12 


0245 


0133 


.0065 


0043 


30.4 


14 


0285 


0154 


.0075 


0049 


26.4 


16 


0325 


0175 


.0085 


0055 


23-3 


18 


0365 


0196 


.0095 


0061 


20.8 


20 


0405 


0218 




0068 


18.8 


22 


0445 


0239 




0074 


17.2 


24 


0485 


0260 




0080 


15-1 


27 


0545 


0292 




0090 


13-5 


30 


0605 


0324 




0099 




38 




040 lr> 




l^, f 


8l? 


44 
56 


10 

+ 


M 


047 + 

053 q 

060 "^ 


8 




+ 


§ 


+ 







^ 


62 


II 


066 " 1 


It 


!l 


s: 


66 


^ 


070 - 


"^ 


"^ 



The alignment should be absolutely accurate in starting. 
To secure this some engineers resort to the use of two diameters 
— each half the length of the fit — differing by but a few thou- 
sandths of an inch. 

Experience shows that, for same allowances, shrink fits hold 
more firmly than pressure fits * 

III. Stress in Hub. — In regard to the tension in the hub due 
to shrinkage or forced fits, completely satisfactor}^ data are 
lacking. A close approximation to the probable tension in the 
inner layer of hub— the one next th e shaft— may be made 

* For valuable data on fits and fittings, see Am. Mach., Mar. 7, 1907. 



l6o MACHINE DESIGN. 

by considering the hub as a thick cylinder under internal pres- 
sure. Then * 

/i =unit stress in pounds per square inch in inner layer; 
pi = internal pressure in pounds per square inch; 
Ti = internal radius of hub in inches; 
^2 = external radius of hub in inches. 

To determine the probable value of pi we may proceed as 
follows: Knowing the maximum forcing pressure in tons, 

/area of surface of fit X^ XPF\ 
^ \ 2 / ' 

this may be presumed to be the resistance offered by friction 
between the surfaces. Assuming the coefficient of friction 
= o.io, the. total pressure between the surfaces is ten times as 
great as the forcing pressure. Reducing this total pressure to 
pounds, and dividing by the area of surface of fit, gives the 
pressure per square inch, or the value of pi. 

.*. ^,- = io,ooo^XPi^. 

112. Problem. — A hub yj inches in diameter and 8 inches 
long is to be forced on a 5-inch shaft. What allowance should 
be made in difference of diameter of bore and shaft ? What 
will be the necessary forcing pressure ? What will be the tensile 
stress in the inner fibers of the hub? 

From the table the forcing-fit allowance for a 5 -inch nominal 
diameter is 0.0105''=^. 

* Ewing's "Strength of Materials," p. 210. 



MEANS FOR PREVENTING RELATIVE ROTATION. i6i 

The forcing pressure in tons 

(SXttXS) X (0.0105) X91 
= 2 = 59-i3> 

since the table gives us PF = gi for this case. 
The internal pressure 

/>i = 10,000X0.0105X91 =9555 lbs. per square inch, 

and 

J 9555(2-5^ + 3-75^) _ 24,811 lbs. per square inch. 

This is a dangerous value for cast iron and points to a greater 
outside diameter of hub or a smaller fit allowance. The rough 
rule of practice is to make he outside diameter of hub equal 
twice the diameter of the shaft for cast-iron hubs This 
would reduce /i to 17,500 lbs., about. Or, using the original 
hub but only one half the tabular allowance, /i becomes 
12,405 lbs. 



CHAPTER X. 



SLIDING SURFACES. 



113. General Discussion. — So much of the accuracy of 
motion of machines depends on the sliding surfaces that their 
design deserves the most careful attention. The perfection of 
the cross-sectional outline of the cylindrical or conical forms 
produced in the lathe depends on the perfection of form of the 
spindle. But the perfection of the outlines of a section through 
the axis depends on the accuracy of the sliding surfaces. All 
of the surfaces produced by planers, and most of those pro- 
duced by milling-machines, are dependent for accuracy on the 
sliding surfaces in the machine. 




Fig. I 10. 



F Q 

Fig. III. 



114. Proportions Dictated by Conditions of Wear. — Suppose 
that the short block A^ Fig. no, is the shder of a slider-crank 
chain, and that it slides on a relatively long guide B. The 
direction of rotation of the crank a is as indicated by the arrow. 
B and C are the extreme positions of the slider. The pressure 
between the slider and the guide is greatest at the mid-position, 
A ; and at the extreme positions, B and C, it is only the pressure 
due to the weight of the slider. Also the velocity is a maximum 

162 



SLIDING SURFACES. 



163 



when the slider is in its mid-position, and decreases towards the 
ends, becoming zero when the crank a is on its center. The 
work of friction is therefore greatest at the middle, and is very 
small near the ends. Therefore the wear would be the greatest 
at the middle, and the guide would wear concave. If now the 
accuracy of a machine's working depends on the perfection 
of ^'s rectilinear motion, that accuracy will be destroyed as the 
guide Z> wears. Suppose a gib, EFG^ to be attached to A, Fig. 
Ill, and to engage with D, as shown, to prevent vertical loose- 
ness between A and D. If this gib be taken up to compensate 
for wear after it has occurred, it will be loose in the middle 
position when it is tight at the ends, because of the unequal 
wear. Suppose that A and D are made of equal length, as in 
Fig. 112. Then when A is in the mid-position corresponding 




Fig. 112. 

to maximum pressure, velocity, and wear, it is in contact with 
D throughout its entire surface, and the wear is therefore the 
same in all parts of the surface. The slider retains its accuracy 
of rectilinear motion regardless of the amount of wear; the 
gib may be set up, and will be equally tight in all positions. 



ra. 



J=ggj^ 



Fig. 113. 



If A and B, Fig. 113, are the extreme positions of a slider, 
D being the guide, a shoulder would be finally worn at C It 
would be better to cut away the material of the guide, as shown 



1 64 MACHINE DESIGN. 

by the dotted line. Slides should always "wipe over" the ends 
of the guide when it is possible. Sometimes it is necessary 
to vary the length of stroke of a slider, and also to change its 
position relatively to the guide. Examples: "Cutter-bars" 
of slotting- and shaping-machines. In some of these positions 
there will be a tendency, therefore, to wear shoulders in the 
guide and also in the cutter-bar itself. This difficulty is over- 
come if the slide and guide are made of equal length, and the 
design is such that when it is necessary to change the position 
of the cutter-bar that is attached to the slide, the position of 
the guide may be also changed so that the relative position of 
slide and guide remains the same. The slider surface will then 
just completely cover the surface of the guide in the mid-position, 
and the slider will wipe over each end of the guide whatever 
the length of the stroke. 

In many cases it is impossible to make the slider and guide 
t)f equal length. Thus a lathe-carriage cannot be as long as the 
bed, a planer-table cannot be as long as the planer-bed, nor a 
planer-saddle as long as the cross-head. When these condi- 
tions exist especial care should be given to the following: 

I. The bearing surface should be made so large in pro- 
portion to the pressure to be sustained that the maintenance; 
of lubrication shall be insured under all conditions. 

II. The parts which carry the wearing surfaces should be 
made so rigid that there shall be no possibility of the localiza- 
tion of pressure from yielding. 

115. Form of Guides. — As to form, guides may be divided 
into two classes: angular guides and flat guides. Fig. 114, a, 
shows an angular guide, the pressure being applied as shown. 
The advantage of this form is, that as the rubbing surfaces 
wear, the slide follows down and takes up both the vertical 
and lateral wear. The objection to this form is that the pres- 
sure is not applied at right angles to the wearing surfaces, as 



SLIDING SURFACES. 



165 



it is in the flat guide shown in h. But in 5, a gib must be pro- 
vided to take up the lateral wear. The gib is cither a wedge 




or a strip with parallel sides backed up by screws. Guides 
of these forms are used for planer-tables. The weight of the 
table itself holds the surfaces in contact, and if the table is light 
the tendency of a heavy side cut would be to force the table 
up one of the angular surfaces away from the other. If the 
table is very heavy, however, there is little danger of this, and 
hence the angular guides of large planers are much flatter 
than those of smaller ones. In some cases one of the guides 
of a planer-table is angular and the other is flat. The side 
bearings of the flat guide may then be omitted, as the lateral 
wear is taken up by the angular guide. This arrangement is 
undoubtedly good if both guides wear down equally fast. 



D 


-7 




] 


.^— =X\c 


/ 


















'^r 






^-\ 




^ 






B 







Fig. 115. 

Fig. 115 shows three forms of sliding surfaces such as are 
used for the cross-slide of lathes, the vertical slide of shapers, 
the table slide of milling-machines, etc. ^ is a taper gib that 
is forced in by a screw at D to take up wear. When it is neces- 
sary to take up wear at B, the screw may be loosened and a 
shim or liner may be inserted between the surfaces at a. C is 
a thin gib and the wear is taken up by means of several screws 



1 66 MACHINE DESIGN. 

like the one shown. This form is not so satisfactory as the 
wedge gib, as the bearing is chiefly under the points of the 
screws, the gib being thin and yielding, whereas in the wedge 
there is complete contact between the metallic surfaces. 

The sliding surfaces thus far considered have to be designed 
so that there will be no lost motion while they are moving, 
because they are required to move while the machine is in 
operation. The gibs have to be carefully designed and accu- 
rately set so that the moving part shall be just " tight and 
loose "; i.e., so that it shall be free to move, without lost motion 
to interfere with the accurate action of the machine. There 
is, however, another class of sliding parts, like the sliding- 
head of a drill-press, or the tailstock of a lathe, that are never 
required to move while the machine is in operation. It is only 
required that they shall be capable of being fastened accu- 
rately in a required position, their movement being simply to 
readjust them to other conditions of work while the machine 
is at rest. No gib is necessary and no accuracy of motion 
is required. It is simply necessary to insure that their posi- 
ion is accurate when they are clamped for the special work 
to be done. 

1 1 6. Lubrication. — The question of strength rarely enters 
into the determination of the dimensions of sliding surfaces; 
these are determined rather by considerations of minimizing 
wear and maintaining lubrication. As long as a film of oil 
separates the surfaces, wear is reduced to a minimum. The 
allowable pressure between the surfaces without destruction 
of the film of lubricant varies with several conditions. To 
make this clear, suppose a drop of oil to be put into the middle 
of an accurately finished surface plate (i.e., as close an approxi- 
mation to a plane surface as can be produced) ; suppose another 
exactly similar plate to be placed upon it for an instant; the 
oil-drop will be spread out because of the force due to the weight 



SLIDING SURFACES. 167 

of the upper plate. Had the plate been heavier it would have 
been spread out more. If the plate were allowed to remain 
a longer time, the oil would be still further spread out, and 
if its weight and the time were sufficient, the oil would finally 
be squeezed entirely out from between the plates, and the 
metal surfaces would come into contact. The squeezing out 
of the oil is, therefore, a function of the time as well as of 
pressure. 

If the surfaces under pressure move over each other, the 
removal of the oil is facilitated. The greater the velocity of 
movement the more rapidly will the oil be removed, and there- 
fore the squeezing out of the oil is also a function of the velocity 
of the rubbing surjaces. 

117. Allowable Bearing Pressure. — Flat surfaces in m.achines 
are particularly difficult to make perfectly true in the first place, 
and to keep true in the course of operation of the machines. 
If they are distorted ever so slightly the pressure between the 
surfaces becomes concentrated at one small area, and the actual 
pressure per square inch is vastly in excess of the nominal 
pressure. 

In consequence of this and the differences in original truth 
and finish of the surfaces, there is no matter in machine design 
in which practice varies more than in the nominal pressure 
allowed per square inch of bearing area of flat sliding surfaces. 

Unwin* gives the following: 

Table XIV. 

Slipper slide-blocks, marine engines 100 lbs. per square inch 

Stationary-engine slide-blocks 25 to 125 " " " 

Stationary -engine slide-blocks usually. ... 30 to 60 " ' ' " ' ' 

Professor Barr f found American practice to vary as follows : 

* "Machine Design," i4Lh ed.. Vol. I, p. 198. 
t Trans. A. S. M. E., Vol. XVIII, p. 753. 



1 68 MACHINE DESIGN. 

Cross-head shoes of high-speed engines: 

Minimum pressure per square inch lO . 5 lbs. 

Maximum pressure per square inch 38 * * 

Mean pressure per square inch 27 * * 

Cross-head shoes of low -speed engines: 

Minimum pressure per square inch 29 lbs. 

Maximum pressure per square inch 58 " 

Mean pressure per square inch 40 " 

In all cases the mean sliding velocity was probably in the 
neighborhood of 600 feet per minute with a maximum velocity 
at the middle of the stroke of about 950 feet per minute. In 
*' low-speed " engines the maximum velocity is only reached 
about one third as many times per minute as in " high-speed " 
engines, although they may have the same mean velocity, and 
it is therefore proper to allow a higher unit value of pressure 
for the former than for the latter. For well-made surfaces 
the maximum values given by Professor Barr may be safely 
used. 

For lower mean speeds than 600 feet per minute they may 
be increased, and for higher speeds decreased, according to some 
law such as 

^7 = 36,000, 

in which formula ^=pressure per square inch, and F = velocity 
of rubbing in feet per minute. 

118. Maintenance of Lubrication. — Regarding the materials 
to be used, brass, bronze, or babbitt metal will run well with 
iron or steel. 

To maintain lubrication a constant flow of oil from a cup is 
desirable. The moving surface should, if possible, have chan- 
nels cut in its face to conduct the oil from the central oil-hole to 
all parts of the surface, as shown in Fig. 116. The oil should 
be forced in where the pressures are heavy. 

Oil-pads may be used as shown in Fig. 117. The shaded 



SLIDING SURFACES. 



i6j 



areas represent porous pads whose lower surfaces just touch the 
surface to be lubricated, and which are kept soaked with oil. 






i 




1 
1 



Fig. ii6. 



Fig. 117. 



For oiling the ways of planer-tables it is customary to use 
rollers placed in oil-filled pockets in the guides. The top of 





Fig. 118. 



the roller is held against the surface of the way by means of 
springs. (See Fig. 118.) 



CHAPTER XI. 

AXLES, SHAFTS, AND SPINDLES. 

119. By Axles, Shafts, or Spindles we denote those rotating 
or oscillating members of machines whose motion is constrained 
by turning pairs. Axle is the name given to such a member 
when it is subjected to a load which produces a bending moment, 
and the only torsional stress is that due to friction. 

When rotating members are subjected chiefly to torsional 
stress, or combined torsion and bending, they are called shafts 
or spindles The former term is used where the part has as 
its function the transmission of energy of rotation from one 
point to another. Examples are liae-shafts and crank-shafts. 

The term spindle, on the other hand, is restricted to those 
rotating members which are directly connected with the tool or 
work and give it an accurate rotative motion. They generally 
fo. m the main axis of the machine. Examples are the lathe and 
drill-spindles. 

120. Axle Design. — The question of axle design will be taken 



■' p '^ 

R ■ Q s 



Fig. 119. 

up first, and the torsional moment due to friction will be neg- 
lected. 

A typical case is shown in Fig. 119. Here the two ends are 

170 



AXLES, SHAFTS, AND SPINDLES. 



171 



purposely not symmetrical. Given the loads P and Q, solution 
is first made for the reactions R and 5 by the ordinary methods 
of mechanics, graphical or analytical. 

The graphical method is best since it gives the moments at 
all sections. Lay off the line M-N, Fig. 120, whose length 
equals I' +l-\-l'', the distance between the points of application 
of R and S. Denote the points of application of R, P, Q, and 
5 by a, h, c, and d, respectively. At h erect a perpendicular and 



r-^ 



■Bli 




Fig. 120. 



lay off on it a vector representing the value of P in pounds. K 
c erect a perpendicular and lay oft' on it a vector representing 
Q on the same scale. At d drop a perpendicular and lay off 
de equal to vector (J, and ej equal to vector P. Select any 
point O as pole, and draw Od, Oe, and Of. Denotb hy g the point 
where Od intersects a perpendicular dropped ftom c, and draw 
from g a parallel to Oe until it intersects a perpendicular dropped 
from b Sit h. From h draw a parallel to Of until it intersects a 
perpendicular dropped from a at j. Dtp vv jd, and parallel to jd 
draw a line through O. This line cuts ^he perpendicular dropped 
from d at the point k. Then vector jk^R, and kd^S, on 
the same scale as was originally used for P and Q, lvalues 
of R and S in pounds are therefore determined. The shaded 
area dghjd is the moment diagram. The vertical ordinates 



172 MACHINE DESIGN. 

included between its bounding lines are proportional to the 
moments at the corresponding points. 

The scale of the moment diagram can be readily determined 
by solving for the actual moment for one section. Select the sec- 
tion at h. The moment Mh is represented by mh and has a 
value =i?/', R being expressed in pounds and /' in inches; the 
value of the moment in inch-pounds can be determined at any 
point, since the scale used is mh inches equals RV inch-pounds. 

For a circular section we have the elastic moment 



M 



hi h^r^ 



c 4 

Mh is the bending moment in inch-pounds; 
jt is the unit stress in outer fiber in pounds per square inch ; 
/ is the plane moment of inertia of the section in biquad- 
ratic inches; 
c is the distance from neutral axis to outermost fiber in 
inches. 
Equating this to the various selected values of Mh and solv- 
ing for r gives the radius of the axle at any point. 

In designing axles, great care must be taken that all forces 
acting are being considered, and that the maximum value of 
each is selected.* 

Thus it has been found that the force due to vertical oscilla- 
tion caused by jar in running is about 40 per cent of the static 
load for car-axles. The axles would therefore have to be de- 
signed for a load 1.4 tim-es the static load. In addition to this 
there is in car-axles a bending moment due to curves, switches, 
and wind-pressures. This may amount as a maximum to the 
equivalent of a horizontal force iJ, equal to 40 per cent of the 
static load, applied at a height of 6 feet from the rail. 

* See further Proc. Master Car-Builders' Assn., 1896; Report of Committee 
on Axles, etc. Also Strength of Railway-Car Axles, Trans. A. S. M. E., 1895. 
Reuleaux, " The Constructor," trans, by H. H. Supplee, Philadelphia, 1893; 
Railway Machinery, Mar. 1907. 



/iXLES, SHAFTS, AND SPINDLES. 173 

When such a careful analysis of the forces has been made, 
jt may be taken for good material, equal to one fourth of the 
ultimate strength, this being a safe value for cases like this, 
where the fibers are subjected to alternate tension and com- 
pression as determined by Wohler and others. 

Had the static load alone been considered in the calculations, 
ft should not have been taken greater than one tenth of the 
ultimate strength. 

121. Shafting Subject to Simple Torsion. — If a short shaft 
is subjected to simple torsion, its diameter may be deter- 
mined very readily by the simple formula for torsional 
moment, 

C 

Here Mt = PR is the torsional moment, P being the force tending 
to twist the piece in pounds and R being the lever-arm of P 
about the axis of the piece in inches. 

/ is the polar moment of inertia of the cross-section of the 

member in biquadratic inches; 
c is the distance from the neutral axis to the outermost fiber 

in inches ; 
/« is the allowable unit stress in pounds per square inch. 
For a solid circular section 



J 

c 


2 


Mr- 


/a7rf3 



and 



which can be solved readily for r, the radius of the shaft. 
For a hollow circular {i.e., ring-shaped) section. 



174 



MACHINE DESIGN, 
J 7:(ri^-r2^) 



2ri 



and 



Mt = 



}s7r(r,^-r2^) 



2r\ 



Here ri is the radius of the outside of the shaft and ^2 is the 
radius of the bore, both in inches. 

The way to solve this is to take r2 as a decimal part of ri. 
Thus, let r2 = bri. It then becomes an easy matter to solve 
for ri. 

122. Shafting Subject to Combined Torsion and Bending. — 
In most cases shafts are subjected to combined torsion and 
bending. Consider the crank-shaft shown in Fig. 121 in side 
and end view. 



n ° ' 




B 


Xi 


k J^^ 












p 




p 




k- 


— I 




Fig. 121. 



B is the center of the bearing, C is the center of the crank- 
pin. At B we have the shaft subjected to a bending nioment, 
Mb=Pl, and also to a twisting moment, Mt=PR. 

Let Meb represent the bending moment which would produce 
the same stress in the outer fiber as Mb and Mt combined. It 
will be called the equivalent bending moment. Then it has 
been found * that 



M. 



eb 



0.35^6 + 0.65V il/52 + M,2 (i) 



* This corresponds to most recent investigations. See Bach, "Elasticitat und 
Festigkeit." For simplicity's sake Bach's coefficient a^ has been dropped, as 
it modifies the result very slightly for wrought iron and steel. 



AXLES, SHAFTS, AND SPINDLES. 1 75 

Also, Meb = — (2) 

c 

For a circular section (2) becomes 
and substitution in (i) gives 



o.35M, + o.65VM,2 + M,2, .... (3) 
4 

which can readily be solved for r, f being given a value equal 
to the maximum allowable unit tensile stress for the material 
and conditions. 

For a hollow circular section 

^''^'"'^~'''^ =o.35Af,+o.65VM;^TM?. . . . (4) 

To solve this express rz as a decimal part of ri. Substitute and 
solve for ri. 

If there are several forces acting, as there are apt to be, the 
method is as follows: First, find Mb due to all the bending 
forces combined. Second, find Mt due to all the twisting forces 
combined. Third, use these values of Mb and Mt in equations 
(i), (3), and (4). Among the forces acting we must not fail 
to include the weight of the shaft and attached parts. 

123. Comparison of Solid and Hollow Shafts. — It is evident 
from (3) and (4) that the dimensions of a solid shaft and a 
hollow shaft of equal strength will have the relationship 



.=^^^ 



r2' 



ri 



176 MACHINE DESIGN. 

If ^2 = 0.6^1, we have 



f = ^o.87ri3, 

^ = o-955''i- 

Hence a hollow shaft whose internal bore is 0.6 of its ex- 
ternal diameter, in order to have the same strength as a solid 
shaft must have its external diameter 1.047 times the diameter 
of the solid shaft. The weight of the hollow shaft will be 70 
per cent of that of the solid shaft. It is obvious that a con- 
siderable saving in weight may be effected without appreciable 
increase in size if the hollow section is adopted. By using nickel 
steel in connection with the hollow section we can get com- 
bined maximum strength and lightness.* 

124. Angular Distortion. — The angle by which a shaft sub- 
jected to torsion is twisted is often an important matter. Let 
this angle be represented by d. Then 

MtLiSo ' 

^^-laT' • V (5) 

i9= angle of torsion in degrees; 
ikff = twisting moment in inch-pounds; 
L = length of shaft in inches ; 

/ = polar moment of inertia in biquadratic inches; 

modulus of elasticity 



G = modulus of torsion = 



2.6 



125. Combined Thrust and Torsion. — When a shaft is sub- 
jected to combined thrust and torsion, the following formula 
has been developed by Prof. A. G. Greenhilhf 

* See "Nickel Steel," a paper by D. H. Browne in Vol. 29 of the Trans. 
A. I M. E. 

t See Proc. of Inst, of M. E., 1883, p. 182. 



AXLES, SHAFTS, AND SPINDLES. i77 

/ = the length of shaft between bearings in iaches; 
P = the end thrust in pounds ; 
£ = modulus of elasticity; 

/ = plane moment of inertia of the section in biquadratic 
inches ; 
If f = twisting moment in inch-pounds. 

This formula is strictly only applicable to vertical shafts, as it 
i^ores the important item of bending due to the weight of the 
shaft and attached parts. 

126. Line-shafts. — Line-shafts are long shafts used to trans- 
mit power. They are made of lengths coupled together and 
supported by bearings at suitable intervals. Pulleys or gears 
are keyed to them, and should always be placed as close to the 
supporting bearings as possible. 

Consider first a length of such a shaft which is subjected to 
pure torsion only, no pulleys being mounted on it. 

Because of its weight, the length of shaft between a pair of 
bearings will sag so that its axis will not be a straight line. 

When a shaft revolves at a high speed, its own inertia gives 
it a tendency to instability independent of the torsion to which 
the shaft is subjected. This is due to the action of centrifugal 
force. 

The "sag" of the shaft causes the center of mass to lie off 
the axis of rotation. At a certain speed the centrifugal force is 
just sufficient to keep the shaft bent. As this critical speed is 
passed the rapidly increasing centrifugal force exceeds the elastic 
forces of the material and the bending becomes large. The 
shaft is then said to ''whirl. " 

The relation existing between the speed, size of shaft, and 



178 MACHINE DESIGN. 

distance between bearings which gives rise to whirling in wrought 
iron or steel shafts is as follows : 



Z = 4oo 



Nl^' 



This applies to an unloaded shaft with bearings which do not 
fix the direction of the ends. 

L = distance between bearings in inches ; 

r = radius of shaft in inches ; 

« = revolutions of shaft per second. 
This becomes a matter of importance in rapidly rotating shafts. 

When there are pulleys on the shaft the value of L naturally 
becomes smaller.* 

If line-shafts are designed wholly for strength, i.e., if d is 
determined by means of equation (3), owing to their length, 
there is apt to be an excessive angular distortion -&. It is 
therefore customary to design them for stiffness and check for 
strength afterwards. 

d should not exceed 0.075° per foot of shaft. 

Combining this rule with the formula (5) for angular dis- 
tortion, 

MtLiSo 
^~ JGt: '' 

for a round wrought-iron or steel solid shaft we get 

J = 4.8<|-^,t 



* For a very complete discussion of " Whirling and Vibration of Shafting," 
see an article by Professor Dunkerly in the Phil. Trans, of the Roy. Soc. of 
London, Vol. 185^, Part I, p. 279. 

t Counting on the fact that the average load is less than the maximum, makers 



of steel shafting use values as small as d 



4 l^-P- 



AXLES, SHAFTS, AND SPINDLES. 179 

when J = diameter of shaft in inches; 

H.P.= horse-power to be transmitted; 
AT" = revolutions per minute of shaft. 

Having determined the diameter which will give sufficient 
stiffness against torsion the allowable distance between sup- 
porting bearings must be calculated. The rule of practice is 
to limit the deflection to i/ioo of an inch to a foot of length. 

Consider first a bare shaft. There are three cases : 

ist. Both ends of the shaft are free to take any direction. 

2d. One end is free and one fixed. 

3d. Both ends are fixed. 

In each case 

L = length of span in inches ; 
t£^= weight of shaft per inch; 
3; = maximum deflection; 

-y- = average deflection per foot of length 

= 1/100 inch; 

L 

•*• y 



2400 



Each case is that of a uniformly loaded beam with a load = ?<;£. 

For case I, the deflection y= — frr. 

77EI 

wL^ 
For case II, the deflection y = ^^. 

-^ ig2 EI 

wL^ 
For case III, the deflection y= ^ ^^ . 

Smce y= , and for round shaftmg I = -r~ and w= 

-^ 2400 ^ 64 

.28 — , while £ = 30,000,000, it follows that 
4 



i8o 



(Case I) 




(Case ID 


L=ioo^^; 


(Case TIT^. 


Z = i3o^^. 



When there are loads due to belt pull, etc., the deflection 
must be determined in each case. For ordmary purposes, 
with the average number of pulleys and amount of belt pull, 
it is safe to take for loaded shafts -fo of the value of L deter- 
mined for bare shafting for the same conditions. Case II 
corresponds most closely to ordinary conditions. 



CHAPTER XII. 

JOURNALS, BEARINGS, AND LUBRICATION.* 

127. General Discussion of Journals and Bearings. — Jour- 
nals and the bearings or boxes with which they engage are 
the elements used to constrain motion of rotation or vibration 
about axes in machines. Journals are usually cylindrical, but 
may be conical, or, in rare cases, spherical. The design of 
journals, as far as size is concerned, is dictated by one or 
more of the four following considerations. 

(i) To provide for safety against rupture or excessive yield- 
ing under the applied forces. 

(2) To provide for maintenance of form. 

(3) To provide against the squeezing out of the lubricant. 

(4) To provide against overheating. 

To illustrate (i), let Fig. 122 represent a pulley on the 
end of an overhanging shaft driven by a belt, ABC. Rota- 
tion is as indicated by the arrow, and the belt tensions are Ti 
and T'2. The journal, J, engages with a box or bearing, D. 
The following stresses are induced in the journal: ToRSiON, 
measured by the torsional moment {T\ — T^r. Flexure, 
measured by the bending moment {Tx"\-T'^a. This assumes 
a rigid shaft or a self-adjusting box. Shear, resuUing from 
the force Ti-{-T2' This journal must therefore be so designed 
that rupture or undue yieldmg shall not result from these 
stresses. 

To illustrate (2), consider the spindle journals of a grind- 

ing-lathe The forces applied are very small, but the form 

of the journals must be maintained to insure accuracy in the 

product of the machine. A relatively large wearing surface 

* See funhcr Vol. 27 Trans. A. S. M. E., p. 420-505. 

181 



l82 



MACHINE DESIGN. 



is therefore necessary, and careful provision must be mad< 
to exclude dust and grit. 




>T, 



F 
, cn 


C 


-^T 




\ \ 

1 T J T } 






1 THT,^ 





Fig. 122. 



To illustrate (3), the pressure upon a journal resulting from 
the applied forces may be sufficiently great to squeeze out the 
lubricant. Metallic contact, heating, and abrasion of the sur- 
faces would result. In what follows, the area of a journal means 
its PROJECTED area; i.e., its length multiplied by its diameter. 

128. Allowable Bearing Pressure. — The allowable pressure 
per square inch of area of a journal varies with several condi- 
tions. The illustration of the drop of oil between two sur- 
face plates, given in the discussion of sliding surfaces (p. 166), 
applies here also. The squeezing out of the oil from between 
the rubbing surfaces of a journal and its box is, therefore, 
a function of the time as well as of pressure. If the sur- 
faces under pressure move over each other, the removal of 
the oil is facilitated. The greater the velocity of movement, 
the more rapidly will the oil be removed, and therefore the 
squeezing out of the oil is also a function of the velocity of 

RUBBING SURFACES. 



JOURNALS, BEARINGS, AND LUBRICATION. 183 

When a journal is subjected to continuous pressure in one 
direction, as, for instance, in a shaft with a constant belt pull, 
or with a heavy fly-wheel upon it, this pressure has sufficient 
time to act, and is therefore effective for the removal of oil. 
But if the direction of the pressure is periodically reversed, as 
in the crank-pin of a steam-engine, the time of action is less, the 
tendency to remove the oil is reduced, and the oil has opportu- 
nity to return between the surfaces. Hence a higher pres- 
sure per square inch of journal would be allowable in the second 
case than in the first. 

If the direction of motion is also reversed, as in the cross- 
head pin of a steam-engine, the oil has not only an opportu- 
nity to return between the surfaces, but is assisted in doing 
so by the reversed motion. Therefore a still higher pressure 
per square inch of journal is allowable. Practical experi- 
ence bears out these conclusions. 

The allowable pressure depends also upon the workmanship 
as shown in the fit of journal and box and the condition of the 
bearing surfaces. The value to be used in each case must be 
decided by the judgment of the designer. The following table 
based upon practice may be taken as a guide: 

Table XV. — Allowable Journal Pressures. 

Pressure in Lbs. 
Kind of Bearing. per Sq. In. of 

Projected Area. 
Bearings for slow speed and intermittent load, such as 

crank-pins of shearing-machines 2000-3000* 

Main journals, center -crank high-speed engines 180- 240 

Jvlain journals, side-crank low-speed engines 160- 220 

Crank -pins of high-speed engines 250- 600 

Crank-pins of low-spe-^d engines 870-1550 

Cross-head pins of high-speed engines 910-1675 

Cross-head pins of low-speed engines 1000-1860 

Car-axle journals 300- 600 

- In \'ol. 27. Trans. A.S.M.E. pp. 496-497, Mr. Oberlin Smith gives examples 
of journal pressures in presses running as high as 20,000 pounds per square inch 
on hardened steel toggle pins; and 7000 pounds per square inch, at a surface speed 
of 140 feel per minute, against the cast iron pitman driving the ram. The journa' 
pressure of the main shaft of the second press was 2400 pounds per square inch. 



1 84 MACHINE DESIGN. 

129. Heating of Journals. — To illustrate (4), even if the 
conditions are such that the lubricant is retained between the 
rubbing surfaces, heating may occur. There is always a fric- 
tional resistance at the surface of the journal; this resistance 
may be reduced (a) by insuring accuracy of form and perfec- 
tion of surface in the journal and its bearings; (b) by insuring 
that the journal and its bearings are in contact, except for the 
filin of oil, throughout their entire surface, by means of rigidity 
of framing or self-adjusting boxes, as the case may demand; 
{c) by selecting a suitable lubricant to meet the conditions and 
maintaining the supply to the bearing surfaces. By these 
means the friction may be reduced to a very low value, but it 
cannot be reduced to zero. 

There must be some frictional resistance, and it is always 
converting mechanical energy into heat. This heat raises the 
temperature of the journal and its bearing. If the heat thus 
generated is conducted and radiated away as fast as it is gener- 
ated, the box remains at a constant low temperature. If, how- 
ever, the heat is generated faster than it can be disposed of, 
the temperature of the box rises till its capacity to radiate heat 
is increased by the increased difference of temperature of the 
box and the surrounding air, so that it is able to dispose of 
the heat as fast as it is generated. This temperature, necessary 
to establish the equilibrium of heat generation and disposal, 
may under certain conditions be high enough to destroy the 
lubricant or even to melt out a babbitt-metal box-lining. Sup- 
pose now that a journal is running under certain conditions 
of pressure and surface velocity, and that it remains entirely 
cool. Suppose next that, while all other conditions are kept 
exactly the same, the velocity is increased All modern experi- 
m-ents on the friction in journals show that the coefficient of 
friction increases with the increase of velocity of rubbing sur- 
face (at speeds above 100 feet per minute). Therefore the in- 



JOURNALS, BEARINGS, AND LUBRICATION. 185 

crease in velocity would increase the frictional resistance at the 
surface of the journal, and the space through which this resist- 
ance acts would be greater in proportion to the increase in 
velocity. The work of the friction at the surface of the journal 
is therefore increased because both the force and the space 
factors are increased. It is this work of friction which has 
been so increased, that produces the heat which tends to raise 
the temperature of the journal and its box. The rate of gen- 
eration of heat has therefore been increased by the increase in 
velocity, but the box has not been changed in any way, and 
therefore its capacity for disposing of heat is the same as it 
was before, and hence the tendency of the journal and its bear- 
ing to heat is greater than it was before the increase in. velocity. 
Some change in the proportions of the journal must be made 
in order to keep the tendency to heat the same as it was before 
the increase in velocity. If the diameter of the journal be 
increased, the radiating surface of the box will be proportion- 
ately increased. But the space factor of the friction will be 
increased in the same proportion, and therefore it will be appar- 
ent that this change has not affected the relation of the rate 
of generation of heat to the disposal of it. But if the length 
of the journal be increased, the work of friction is the same 
as before and the radiating surface of the box is increased and 
the tendency of the box to heat is reduced. If, therefore, the 
conditions are such that the tendency to heat in a journal, 
because of the work of the friction at its surface, is the vital 
point in design, it will be clear that the length of the journal 
is dictated by it, but not the diameter. The reason why high- 
speed journals have greater length in proportion to their diam- 
eter than low-speed journals will now be apparent. 

The lost work per minute due to friction may be expressed 
by PfjizdN, in which P=mean total pressure on journal, 
// = coefficient of friction, J = diameter of journal, and N =revo- 



i86 



MACHINE DESIGN. 



lutions per minute. This energy is all converted into heat 
which should be dissipated through a surface which is pro- 
portional to the projected area of the journal, dl. It follows 
that, other conditions remaining constant, the projected area 
should be proportional to the heat generated, and we may write 



PfiTtdN 
dl 

\ 1 = 



= K. 



PjmN_ 
K 



K must be experimentally determined for a given set of condi- 
tions. Sufficient data are not available to form a general table 
of values oi K. 

130. Journal Proportions. — The proportions of engine-jour- 
nals * may be seen in the following table : 

Table XVI. 



Kind of Journal. 


Value of J. 




Minimum. 


Maximum. 


Average. 


Main journal, center-crank high-speed engine.. . . 

Main journal, side-crank low-speed engine 

Crank -pin high-speed engine 

Crank-pin low-speed engine 

Cross-head pin high-speed engine 


2.0 
1-7 

I.O 
I.O 


3-0 
2. I 

2.0 

1-5 


2.2 
1.9 
I. of 
I. if 
1.25 
1-3 


Cross-head pin low-speed engine 





131. Materials to be Used. — Regarding the materials of 
journals and their boxes the following general statements may 
be made. It must be borne in mind that the terms babbitt, 
brass, and bronze cover wide ranges of alloys of varying values. 

Cast iron, wrought iron, soft steel, and hard steel will all run 



* Professor Barr on "Current Practice in Engine Proportions," Trans. A. S. 
M. E.. Vol. 18, p. 737. 

t Unpublished data, same investigations. 



JOURNALS, BEARINGS, AND LUBRICATION. 187 

well at almost any speed on babbitt metal. The pressure per 
square inch which an ordinary babbitt bearing will stand when 
running cool {i.e., at very slow speed), before being squeezed 
out, has been found to be something over 2000 lbs.* 

Cast iron, wrought iron, soft steel, and hard steel will all 
run well on brass and bronze. Brass and bronze of ordinary 
compositions will carry 5000 lbs. per square inch without suffer- 
ing destruction. Bronze, however, is much better than brass. 

Cast iron will run on cast iron where, owing to large bear- 
ing surfaces, the unit pressure is light. Where the pressure and 
speed are high, as in engine-journals, this will not work.f 

In the same way steel will run on cast iron even at high speeds 
if the pressure is light. It has been found that steel will not 
run on cast iron in engine- journals. J 

Wrought iron, soft steel, and hard steel will all run on hard 
steel. 

Steel under steel if hardened and polished will run under as 
high a pressure as 50,000 lbs. per square inch. 

132. Calculation of Journals for Strength. — Journals gener- 
ally form parts of axles on shafts, and the calculation of their 
diameter for strength becomes part of the calculation of the 
shaft. The principles have been developed at length in the 
preceding chapter and need not be repeated here. 

If the journal is so held that it may be considered as sub- 
jected to pure shearing stress, like the crank-pin of a center- 
crank engine, then 

in which P = total maximum load ; 

A = total area subjected to stress; 

/s=safe shearing stress for the conditions. 

* C. F. Porter, Trans. A. S. M. E., Vol. Ill, p. 227. 
t Trans. A. S. M. E., Vol. VI, pp. 853-854. 
t Ibid. 



1 88 MACHINE DESIGN. 

For a journal subjected to a pure bending moment, 

which becomes Pl= for a solid circular shaft. P/ = bendinj:c 

4 \ "^ 

moment, / = safe unit stress, and r = radius of shaft. This can 

readily be solved for r. 

If the journal be hollow, 

An ' 

fi being the external and f2 the internal diameter. 

For combined bending and twisting such as the main journal 
of a side-crank engine is subjected to, the expression for a solid 
journal is 



.Ttr 



/— =o.35il/, +0.65^^,2 +M?. 
For a hollow circular section 

^^^^^^^^ =0.35^, +o.65Vif7T¥?' 

M^ being the bending moment and Mt the twisting moment. 

In general it will be found that journals proportioned for 
strength merely will not have sufficient area to prevent heating, 
so this item must not be overlooked. 

133. Problem. — Design the main journal of a side-crank 
low-speed engine. 
Diameter of cylinder = 16 ins. 
Length of stroke ^36 ins. 

Net forward pressure = 100 lbs. per square inch of piston area. 

Suppose the engine capable of carrying full pressure to half- 
stroke. 



JOURNALS, BEARINGS, AND LUBRICATION. 



189 



The area ot piston =201.06 square inches. 

.*. total net forward pressure = 20,106 lbs. 

At point of maximum torsional effect, which corresponds to 
the position of maximum velocity of piston, no energy is used 
in accelerating reciprocating parts, and 

FpVp = FcVc; 

Fp =net forward force on piston; 
Vp = velocity of piston; 
jPc= force on crank; 
Vc = velocity of crank. 

Since v^ is less than Vp for this position, F^ is greater than 

F,, since F^=^. 

Assuming a connecting-rod length equal to five and a half 
crank lengths gives (Appendix) i^c = 20, 500 lbs. 

Since the crank length is 18 inches, and at this position the 
crank and connecting-rod are nearly at a right angle with 
each other, there is a twisting moment at the journal equal to 

Mt = 20,500 X 18 = 369,000 inch-lbs. 

There is also a bending moment equal to 20,500 X the dis- 
tance from center of crank-pin to center of main journal. In 

most cases this distance must be 
assumed; for, although the length 
of the crank-pin and the thickness 
of the crank may be known, the 
length of the main journal is un- 
Y known, since this length and the 
journal diameter are the very 
dimensions sought. Assume then 
^^' ^^^' that the crank-pin is 6 inches long, 

the crank 3 inches thick, and the middle of main journal 6 



] 



i<-3->i*-a- 



-6-^ 



190 MACHINE DESIGN. 

inches from the inner face of crank as shown in Fig. 123. 
This will give 12 inches as the lever-arm; .*. the bending moment, 
Mf,, = 20, 500 X 1 2 = 246,000 inch-lbs. 

The equivalent bending moment to the combined actual 
bending and twisting moments 

= M,t 

= 0.35 X 246,000 -f 0.65V 246000^ + 3690002 
= 374,375 inch-lbs. 

But M,,J^', 

4 



.3 — 



4X374375 



For a main shaft like this / may be taken = 12,000 lbs. 
per square inch for steel. 



.-. ,4 



4X374375 



71 X 12000 

= 3.41 inches; 

.*. diameter of journal = 2X3.41 =6.82, say 7 inches. 

The length according to practice would be about twice this 
diameter,* or 14 inches. This would give a projected area of 
98 square inches and a pressure of something over 200 lbs. 
per square inch of bearing due to steam-pressure alone. 

To get the actual maximum pressure on the journal it 
would be necessary to know the weight of the shaft, flywheel, 
and other attached parts, and properly combine the pressure 
due to these with the pressure due to the steam. 

The rough rule of practice for Corliss engines is to make 
the diameter of main journal equal to one half the diameter 
of the cylinder. 

* See Table XVI, p. 186. 



JOURNALS, BEARINGS, AND LUBRICATION, 191 

134. Problem. — Design the crank-pin for the same engine. 
It will be found that the crank-pin must be designed with refer- 
ence to maintaining lubrication, and that it will have an excess 
of strength. 

Allowing 1200 lbs. per square inch of area,* and noting 
from the table that the average practice for this type of engine 
is to make the length of the pin = i.iXthe diameter,f it fol- 
lows that 

1200 
but 1=1. id; 

/. 1.1^2= — ^ — 
1200 

and d = 4 inches, nearly; 

.*. /=i.iX4 = 4i inches, say. 

Checking this for strength, considering the pin subjected 

/ 
to a bending moment P-, we write 

P = 20500 lbs., 

-= — = 2.2s inches; 
22 

r=2 inches, 

/ = stress in outer fiber; 

. ^ 4X20500X2.25 „ . ^ 

/. / = —7^ ==7300 lbs. per square mch; 

/^ X o 

which is, of course, a perfectly safe value for wrought iron or 
steel. 

* See Table XV, p. 183. t See Table XVI, p. 186. 



19' 



MACHINE DESIGN. 



135. Problem. — Design the cross-head pin for the same 
engine. This pin also should be designed for maintaining 
lubrication. Allowing 1400 lbs. per square inch as the per- 
missible pressure on the journal,* and noting that the length 
may be taken as 1.3 times the diameter from average practice f 
gives 

1400 
^ 1400 

c? = 3f inches; 
.*. / = 4j inches. 



and 



Checking this for strength it is evident that the only v^ay 




Fig. 124. 



this pin can fail is by shearing on two surfaces, A-B and D-E 
(see Fig. 124). 



/«27rr2; 
20500 



2.84 



= 1150 lbs. per square inch. 



This leaves so great a margin of safety that some manu- 
facturers make the cross-head pin of two parts, an inner pin 
of soft, resilient material, sufficiently large to resist the shear- 
ing stress, and an outer hard-steel bushing which surrounds 
the soft pin, but is not allowed to turn on it. The nature of 



* See Table, XV, p. 183. 



t See Table XVI, p. 186. 



JOURNALS, BEARINGS, AND LUBRICATION. 



193 



the forces acting on a cross-head pin tend to wear it to an oval 
cross-section. As such wear takes place the bushing can readily 
be given a quarter turn and clamped in the new position. 
(See Fig. 124.) 

136. Maintenance of Form. — Journals whose maintenance 
of form is of chief importance must be designed from prece- 
dent, or according to the judgment of the designer. No 
theory can lead to correct proportions. In fact these pro- 
portions are eventually determined by the process of Machine 
Evolution. 

137. Thrust-journals. — When a rotating-machine part is 
subjected to pressure parallel to the axis of rotation, means 
must be provided for the safe resistance of that pressure. In 
the case of vertical shafts the pressure is due to the weight 
of the shaft and its attached parts, as the shafts of turbine 
water-wheels that rotate about vertical axes. In other cases 
the pressure is due to the working force, as the shafts of pro- 
peller-wheels, the spindles of a chucking-lathe, etc. The end- 
thrusts of vertical shafts are very often resisted by the "squared- 
up " end of the shaft. This is inserted in a bronze or brass 
"bush," which embraces it to prevent lateral motion, as in 
Fig. 125. If the pressure be too great, the end of the shaft 
may be enlarged so as to increase the bearing surface, thereby 
reducing the pressure per square inch. This enlargement 



^^"^m 






i 



r — ^<^^ 



Fig. 125. 



Fig. 126. 



must be within narrow limits, however. (See Fig. 126.) AB is 
the axis of rotation, and ACD is the rotating part, its bear- 



194 



MACHINE DESIGN. 



ing being enlarged at CD. Let the conditions of wear be con- 
sidered. The velocity of rubbing surface varies from zero 
at the axis to a maximum at C and D. It has been seen that 
the increase of the velocity of rubbing surface increases both 
the force of the friction and the space through which that 
force acts; it therefore increases the work of the friction, and 
therefore the tendency to wear. From this it will be seen 
that the tendency to wear increases from the center to the 
circumference of this ''radial bearing," and that, after the 
bearing has run for a while, the pressure will be localized near 
the center, and heating and abrasion may result. For this 
reason where there is severe stress to be resisted, the bearing is 
usually divided up into several parts, the result being what is 
known as a ''collar thrust-bearing," as shown in Fig. 127. 




Fig. 127. 

By the increase in the number of collars, the bearing surface 
may be increased without increasing the tendency to unequal 
wear. The radial dimension of the bearing is kept as small 
as is consistent with the other considerations of the design. 
It is found that the " tractrix," the curve of constant tangent, 
gives the same work of friction, and hence the same tendency 
to wear in the direction of the axis of rotation, for all parts of 
the wearing surface. (See " Church's Mechanics," page 181.) 

This has been very incorrectly termed the "anti-friction" 
thrust-bearing. This is far from being the case. The friction 
work for this and all conical thrust-bearings can be shown 



JOURNALS, BEARINGS, AND LUBRICATION. IQS 

readily to be excessive. Their one advantage is that they are 
easily adjustable. In general they are to be avoided. 

The pressure that is allowable per square inch of projected 
area of bearing surface varies in thrust-bearings with several 
conditions, as it does in journals subjected to pressure at right 
angles to the axis.* Thus, in the pivots of turntables, swing- 
bridges, cranes, and the like the movement is slow and never 
continuous, often being reversed; and also the conditions are 
such that ''bath lubrication" may be used, and the allowable 
unit pressure is very high — equal often to 1500 lbs. per square 
inch, and in some cases greatly exceeding that value. The 
following table may be used as an approximate guide in the 
designing of collar-bearings. The material of the thrust-journal 
is wrought iron or steel, and the bearing is of bronze or brass 
(babbitt metal is seldom used for this purpose). 

Table XVILf 

Allowable Unit 
Mean Velocity of Rubbing Surface, Pressure, lbs/in^ of 

Feet per Minute. Projected Area of 

Rubbing Surface. 

Slow and intermittent 1 500 

50 200 

50 to 100 100 

100 to 150 75 

1 50 to 200 60 

Above 200 50 

If the journal is of cast iron and runs on bronze or brass, the 
values of allowable pressure given should be divided by two. 

The most efficient forms of thrust-bearings are those % employ- 
ing the principles shown in Fig. 128. 

* See Proc. Inst. M. E., 1888 and 1891, for reports on experiments with thrust- 
bearings. 

t Reuleaux, "Constructor," p. 65, gives for steel on bronze, 
Slow moving pivots, d=o.o2,$\/P. 
Up to 150 rev. per minute, J=o.o5V'P. 
Above 150 rev. per minute, d=o.oo^\/Pn. 
<i= diameter of shaft in inches, P = total load in pounds, w=revs. per minute. 
These formulas give higher unit pressures than the table derived from the experi- 
ments of the Inst. M.E., which are particularly applicable to collar thrust bearings. 
% See Trans. A. S. M. E., Vol. VI, p. 852, and Proc. Inst. M.E., 1888, p. 184. 



196 



MACHINE DESIGN. 




Between the end of the shaft and the bottom of the step a 
series of accurately finished disks are introduced. The disks are 
alternately hard steel and bronze, the top one is fastened to the 

shaft, the lower to the step, 
and the rest are free. As in- 
dicated, each disk has a hole 
through the middle and radial 
grooves to permit the lubricant 
to have access between the disks. 
The effect of centrifugal force 
when the shaft is rotating is to 
force the oil outward from be- 
tween the plates and upward. 
It is collected in the annular 
chamber a-a and flows from there down the drilled passages back 
to the bottom of the bearing. This is equivalent to a continuous 
automatic pump action supplying oil to the surfaces. This form 
of bearing reduces the relative motion between successive surfaces 
to a minimum, thereby allowing much higher pressures to be car- 
ried. A similar arrangement of loose disks can be used to great 
advantage on small propeller shafts and on worm shafts. 

For thrust-bearings in which the lubricant is automatically 
circulated, or supphed by a force-pump so as to ''float " the jour- 
nal, the allowable unit bearing pressures become quite great, 
examples of satisfactory operation at loads as high as 1000 lbs. per 
sq. in. being known. With a lubricant of suitable viscosity the 
conditions, at sufficiently high speeds, would tend to give practically 
fluid friction, i.e., the frictional resistance would be independent of 
the pressure. As the speed decreases the tendency to maintain the 
oil-film grows less, however, and there are critical speeds corre- 
sponding to certain loads at which the film appears to break down 
and seizing takes place. There are not sufficient data to fully 
tabulate these for various oils and temperatures. Where special 
means of forced lubrication are not employed it will be safe in the 
design of ordinary thrust-journals to use the unit pressures given 
for collar-bearings in Table XVII. 



JOURNALS, BEARINGS, AND LUBRICATION. 197 

When bearings have to be used where corrosion or electro- 
lytic action is to be feared, as in turbine work, glass and the 
end grain of very hard woods have been used successfully as 
bearing materials. 

138. Problem. — It is required to design the collar thrust- 
journal that is to receive the propelling pressure from the screw 
of a small yacht. The necessary data are as follows: The 
maximum power delivered to the shaft is 70 H.P.; pitch of 
screw is 4 feet; slip of screw is 20 per cent; shaft revolves 250 
times per minute; diameter of shaft is 4 inches. 

For every revolution of the screw the yacht moves forward 
a distance = 4 feet less 20 per cent "=3.2 feet, and the speed of 
the yacht in feet per minute = 2 50X3. 2 =800. 

70 H. P. = 70X33,000 = 2,310,000 ft. -lbs. per minute. 

This work may be resolved into its factors of force and space, 
and the propelling force is equal to 2,310,000^800 = 2900 lbs., 
nearly. 

The shaft is 4 inches in diameter, and the collars must project 
beyond its surface. Estimate that the mean diameter of the 
rubbing surface is 4.5 inches, then the mean velocity of rubbing 

surface would equal 4.5 X — X 250 = 294 feet per minute. The 

allowable value of pressure per square inch of journal surface for a 
velocity above 200 feet per minute is 50 lbs. The necessary area 
of the journal surface is therefore = 2900-^-50 = 58 square inches. 
It has been seen that it is desirable to keep the radial dimen- 
sion of the collar surface as small as possible in order to have 
as nearly the same velocity at all parts of the rubbing surface 
as possible. The width of collar in this case will be assumed = 
0.75 inch; then the bearing surface in each collar 

5.52x71 42x7: 

= — = 23.7-12.5 = 11.2 sq. m. 

4 4 

Then the number of collars equals the total required area 
divided by the area of each collar =58-^ 11.2 = 5.18, say 6. 



198 M, 4 CHINE DESIGN. 

139. Bearings and Boxes. — The function of a bearing or box 
is to insure that the journal with which it engages shall have 
an accurate motion of rotation or vibration about the given axis. 
It must therefore fit the journal without lost motion; must 
afford means of taking up the lost motion that results neces- 
sarily from wear ; must resist the forces that come upon it 
through the journal, without undue yielding ; must have the 
wearing surface of such material as will run in contact with 
the material of the journal with the least possible friction and 
least tendency to heating and abrasion; and must usually 
include some device for the maintenance of the lubrication. The 
selection of the materials and the providing of sufficient strength 
and stiffness depends upon principles already considered, and 
so it remains to discuss the means for the taking up of necessary 
wear and for providing lubrication. 

Boxes are sometimes made solid rings or shells, the journal 
being inserted endwise. In this case the wear can only be 
taken up by making the engaging surfaces of the box and journal 
conical, and providing for endwise adjustment either of the 
box itself or of the part carrying the journal. Thus, in Fig. 129, 
the collars for the preventing of end motion while running are 
jam-nuts, and looseness between the journal and box may be 
taken up by moving the journal axially toward the left. 





Fig. 129, Fig. 130. 

By far the greater number of boxes, however, are made in 
sections and the lost motion is taken up by moving one or more 
sections toward the axis of rotation. The tendency to wear is 
usually in one direction, and it is sufficient to divide the box 



JOURNALS, BEARINGS, AND LUBRICATION. 199 

into halves. Thus, in Fig. 130, the journal rotates about the 
axis O, and all the wear is due to the pressure P acting in the 
direction shown. The wear will therefore be at the bottom of 
the box. It will suffice for the taking up of wear to dress ofif 
the surfaces at aa, and thus the box-cap may be drawn further 
down by the bolts, and the lost motion is reduced to an admis- 
sible value. "Liners," or ''shims," which are thin pieces of 
sheet metal, may be inserted between the surfaces of division 
of the box at aa, and may be removed successively for the lower- 
ing of the box-cap as the wear renders it necessary. If the axis 
of the journal must be kept in a constant position, the lower 
half of the box must be capable of being raised. 

Sometimes, as in the case of the box for the main journal of 
a steam-engine shaft, the direction of wear is not constant. 
Thus, in Fig. 131, ^ represents the main shaft of an engine. 
There is a tendency to wear in the direction B, because of the 
weight of the shaft and its attached parts; there is also a ten- 
dency to wear because of the pressure that comes through the 
connecting-rod and crank. The direction of this pressure is 
continually varying, but the average directions on forward and 



>c 





Fig. 132. 

return stroke may be represented by C and D. Provision needs 
to be made, therefore, for the taking up of wear in these two 
directions. If the box be divided on the line EF, wear will be 
taken up vertically and horizontally by reducing the hners. 
Usually, however, in the larger engines the box is divided into 
four sections. A, B, C, and D (Fig. 132), and A and C are 
capable of being moved toward the shaft by means of screws 



200 



MACHINE DESIGN. 



or wedges, while D may be raised by means of the insertion 
of " shims/' 

The lost motion between a journal and its box is sometimes 
taken up by making the box as shown in Fig. 133. The exter- 
nal surface of the box is conical and fits in a conical hole in 
the machine frame. The box is split entirely through at A, 
parallel to the axis, and partly through at B and C The ends 
of the box are threaded, and the nuts E and F are screwed on. 
After the journal has run long enough so that there is an unal- 
lowable amount of lost motion, the nut F is loosened and E 
is screwed up, the effect being to draw the conical box 
further into the conical hole in the machine frame; the hole 




Fig. 133. 

through the box is thereby closed up and lost motion is reduced. 
After this operation the hole cannot be truly cylindrical, and 
if the cylindrical form of the journal has been maintained, it 
will not have a bearing throughout its entire surface. This is 
not usually of very great importance, however, and the form of 
box has the advantage that it holds the axis of the journal in 
a constant position. As far as is possible the box should be 
so designed as to exclude all dust and grit from the bearing 
surfaces. 

All boxes in self-contained machines, like engines or machine 
tools, need to be rigidly supported to prevent the localization 
of pressure, since the parts that carry the journals are made as 
rigid as possible. In line shafts and other parts carrying journals, 
when the length is great in comparison to the lateral dimensions, 



JOURNALS, BEARINGS, AND LUBRICATION. 



201 



some yielding must necessarily occur, and if the bo-xes were 
rigid, localization of pressure would result. Hence ''self- 
adjusting" boxes are used. A point in the axis of rotation at 
the center of the length of the box is held immovable, but the 
box is free to move in any way about this point, and thus adjusts 
itself to any yielding of the shaft. This result is attained as 
shown in Fig. 134. O is the center of the motion of the box; 




Fig. 134. 
B and A are spherical surfaces formed on the box, their center 
being at O. The support for the box contains internal spherical 
surfaces which engage with A and B. Thus the point O is always 
held in a constant position, but the box itself is free to move in 
any way about O as a center. Therefore the box adjusts 
itself within Hmits to any position of the shaft and hence the 
localization of pressure is impossible. 

In thrust-bearings for vertical shafts the weight of the shaft 
and its attached parts serves to hold the rubbing surfaces in 
contact and the lost motion is taken up by the shaft following 
down as wear occurs. In collar thrust-bearings for horizontal 
shafts the design is such that the bearing for each collar is 
separate and adjustable. The pressure on the different collars 
may thus be equalized.* 

140. Lubrication of Journals. f — The best method of lubrica- 
tion is that in which the rubbing surfaces are constantly sub- 

* For complete and varied details of marine thrust-bearings see "Maw's Mod. 
ern Practice in Marine Engineering." 

t See "Lubrication and Lubricants," by Archbutt and Deeiey, London. 



202 M/ICHINE DESIGN. 

merged in a bath of lubricating fluid. This method should be 
employed wherever possible if the pressure and surface velocity 
are high. Unfortunately it cannot be used in the majority of 
cases. It is not necessary that the whole surface be su'> 
merged. If a part of the moving surface runs in the oil-bath it 
is sufficient.* The same result is accomplished by the use of 
chains and rings encircling the journals and dipping into oil- 
pockets, as described later in this section. The effect is to 
form a complete film of oil enveloping the journal. To allow 
this it is evident that the bore of the bearing must be slightly 
greater than the diameter of the journal and a series of ' 'running- 
fit allowances " will be found in Table XIII, § net These 
should be increased with increase of running speeds. 

The oil film may be conceived to be made up of a series of 
layers, the cne next the bearing surface remaining stationary 
with regard to it, while the layer in immediate contact with the 
shaft rotates with the latter. The intermediate layers, therefore, 
slip upon each other as the shaft rotates and the friction becomes 
very closely akin to "fluid friction" with the bearing floating 
upon the lubricant, there being no contact between the metallic 
surfaces. Fig. 134 A shows the conditions of pressure existing 
in the film in Tower's classic experiments. It is impossible 
to introduce oil satisfactorily at the points where the film is 
under pressure; it should be introduced and distributed where 
the pressure is least. Under the action of the load the edges 
of the boxes tend to '' pinch in " and scrape off the film from the 
journal. To prevent this these edges should be cut away, thus 
also forming an excellent oil channel for longitudinal distribution 
of the oil where the pressure is least. An excellent arrange- 
ment of boxes for distributing the oil and maintaining the film 

* Tower's experiments, Proc. Inst. M. E., 1883 and 1885. See further Prof. 
Reynold's paper "On the Theory of Lubrication," Phil. Trans., 1886, Part 1, 
pp. 157-234. 

t Professor Reynolds states, in Phil. Trans. 1886, Part i, p. 161, that if viscosity 
were constant the friction would be inversely proportional to the difference in radii 
of the journal and the bearing. 



204 



MACHINE DESIGN. 



is shown in Fig. 134 B, which is copied from Vol. 27, Trans. 
A. S. M. E., p. 484. With true film lubrication it is quite defi- 
nitely established that the coefficient of friction varies directly 
as the square root of the surface velocity and inversely as the 

C \/y 



unit pressure, so that // 



■y where /i = coefficient of fric- 



tion, c is a constant depending upon the lubricant (being .21 
for rape oil, .29 for lard oil, .32 for mineral oil and .43 for mineral 




TOP HALF Fig. 134 B. bottom half 

grease), v is the velocity of rubbing in feet per second, and p 
is the pressure in pounds per square inch of projected bearing 
area. This holds for rubbing speeds from 75 to 450 feet per 
minute. // varies as Vv at speeds from 500 to 1000 ft., and dots 
not increase with v after a speed of 3500 feet is reached. 

With pad lubrication or where the oil is fed drop by drop there 
is a tendency for the film to be too thin or to break down, allow- 
ing contact of the metallic surfaces, and the highly favorable 
condition of fluid friction disappears. The conditions then lie 
between " fluid friction " and " solid friction " and are too com- 
plex for the statement of consistent resuhs but it may be approxi- 



JOURNALS, BEARINGS, AND LUBRICATION. 



205 



mately stated that, with good pad lubrication, the coefficient 
of friction will be about twice that of film lubrication. With 
drop by drop lubrication the value of the coefficient may become 
anything between twice that for film lubrication (i.e. = .01), 
and 0.18, the value determined by Morin for dry journals. It 
becomes apparent that some system of forced or flooded lubri- 
cation whereby a continuous film is insured is of utmost value in 
maintaining efficiency. 

Let /, Fig. 135, represent a journal with its box, and let A, 
By and C be oil-holes. If oil is introduced into the hole Aj it 



■ 



m 



Fig. 135. 



will tend to flow out from between the rubbing surfaces by the 
shortest way, i.e., it will come out at D. A small amount 
will probably go toward the other end of the box because of 
capillary attraction, but usually none of it will reach the middle 
of the box. If oil be introduced at C, it will come out at E. A 
constant feed, therefore, might be maintained at A and C, and 
yet the middle of the box might run dry. If the oil be introduced 
at Bj however, it tends to flow equally in both directions, and 
the entire journal is lubricated. The conclusion follows that 
oil ought, when possible, to be introduced at the middle of the 
length of a cylindrical journal. It should be introduced as far 
as possible from the side where the forces press the journal and 
box closest together.* If a conical journal runs at a high velocity, 
the oil under the influence of centrifugal force tends to go to 
the large end of the cone, and therefore the oil should be intro- 
duced at the small end to insure its distribution over the entire 
journal surface. 

* Tower's experiments, Proc. Inst. M.E., 1S83 and 1885. 



20 6 



MACHINE DESIGN. 



If the end of a vertical thrust-journal whose outline is a 
cone or tractrix, as in Fig. 136, dips into a bath of oil, J5, the 
oil will be carried by its centrifugal force, if the velocity be 
high, up between the rubbing surfaces, and will be delivered 
into the groove A A. If holes connect A and B, gravity will 
return the oil to B, and a constant circulation will be main- 
tained. If the thrust-journal has simply a fiat end, as in Fig. 
137, the oil should be supplied at the center of the bearing; 
centrifugal force will then distribute it over the entire surface. 
If the oil is forced in under a pressure sufficient to ''float" the 
shaft the friction will be greatly reduced. Vertical shaft thrust- 
journals may usually be arranged to run in an oil-bath. Marine 
collar thrust-journals are always arranged to run in an oil-bath. 




Fig. 136. 



Fig 137. 




zztssz, 



Fig. 138. 



Sometimes a journal is stationary, and the box rotates 
about it, as in the case of a loose pulley, Fig. 138. If the oil 
is introduced into a tube, as is often done, its centrifugal 
force will carry it away from the rubbing surface. But if a 
hole is drilled in the axis of the journal, the lubricant intro- 
duced into it will be carried to the rubbing surfaces as required. 
If a journal is carried in a rotating part at a considerable dis- 
tance from the axis of rotation, and it requires to be oiled while 
in motion, a channel may be provided from the axis of rota- 



JOURNALS, BEARINGS, AND LUBRICATION. 



207 



tion, where oil may be introduced conveniently, to the rub- 
bing surfaces, and the oil will be carried out by centrifugal 
force. Thus Fig. 139 shows an engine-crank in section. Oil 
is introduced at h, and centrifugal force carries it through 
the channel provided to a, where it serves to lubricate the rub- 



Fig. 139. 




Fig. 140. 



bing surfaces of the crank-pin and its box. If a journal is 
carried in a reciprocating machine part, and requires to be 
oiled while in motion, the " wick -and wiper " method is one of 
the best. (See Fig. 140.) An ordinary oil-cup with an adjust- 
able feed is mounted in a proper position opposite the end of 
the stroke of the reciprocating part, and a piece of flat wick 
projects from its delivery-tube. A drop of oil runs down and 
hangs suspended at its end. Another oil cup is attached to 
the reciprocating part, which carries a hooked ^'wiper,^' C. 
The delivery-tube from C leads to the rubbing surfaces to be 
lubricated. When the reciprocating pait reaches the end of 
its stroke the wiper picks off the drop of oil from the wick 
and it runs down into the oil-cup C, and thence to the sur- 
faces to be lubricated. This method applies to the oiling of 
the cross-head pin of a steam-engine. The same method is 
sometimes applied to the crank-pin, but here, through a part 



208 



MACHINE DESIGN. 



of the revolution, the tendency of the centrifugal force is to 
force the oil out of the cup, and therefore the plan of oiling 
from the axis is probably preferable. 

When journals are lubricated by feed-oilers, and are so 
located as not to attract attention if the lubrication should fail 
for any reason, " tallow-boxes " or '' grease-cups " are used. 
These are cup-like depressions usually cast in the box-cap 
and communicating by means of an oil-hole with the rubbing 
surface. These cups are filled with grease that is solid at 
the ordinary temperature of the box, but if there is the least 
rise in temperature because of the failure of the oil- supply, 
the grease melts and runs to the rubbing surfaces, and sup- 
plies the lubrication temporarily. This safety device is used 
very commonly on line-shaft journals. 

The most common forms of feed-oilers are: I. The oil-cup 
with an adjustable valve that controls the rate of flow. 11. The 
oil-cup with a wick feed (Fig. 141). The delivery has a tube 






Fig. 141, 



Fig. 142. 



Fig. 143. 



inserted in it which projects nearly to the top of the cup. In 
this tube a piece of wicking is inserted, and its end dips into 
the oil in the cup. The wick, by capillary attraction, carries 
the oil slowly and continuously over through the tube to the 



JOURNALS, BEARINGS, AND LUBRICATION. 209 

rubbing surfaces. III. The cup with a copper rod (Fig. 142J. 
The oil-cup is filled with grease that melts with a very slight 
elevation of temperature, and ^ is a small copper rod dropped 
into the delivery-tube and resting on the surface of the journal. 
The slight friction between the rod and the journal warms 
the rod and it melts the grease in contact with it, which runs 
dowTi the rod to the rubbing surface. IV. Sometimes a part 
of the surface of the bottom half of the box is cut away and 
a felt pad is inserted, its bottom being in contact with an oil- 
bath. This pad rubs against the surface of the journal, is 
kept constantly soaked with oil, and maintains lubrication. 

Ring-and-chain lubrication may be considered as special 
forms of bath lubrication. Fig. 143 show^s a ring oiling bearing. 
A loose ring rests on top of the journal, the upper box being 
cut aw^ay to permit this; the ring surrounds the lower box 
and extends into a reservoir filled with oil. The rotation of 
the shaft carries the ring with it, which, in turn, brings up a 
constant supply of oil from the reservojr. The annular spaces 
A- A catch all oil which works out along the shaft and return 
it to the reservoir. 

Graphite is winning a deservedly high place as a lubricant for 
certain conditions. Its action is to reduce "solid friction" by 
filling the inequalities in the surfaces of the relatively moving 
members, giving each a smooth, slippery coating, thereby reduc- 
ing the coefficient of friction. It is particularly useful when the 
conditions of pressure or temperature are such as would tend 
to squeeze out, gum, or destroy liquid lubricants, if these were 
used alone. 

Although it may be applied in some cases in dry flake form, 
it is customary to use it in the form of a mixture with oils, grease, 
or even water. Caution must be observed that the graphite used 
is free from all grit. 



CHAPTER XIII. 
ROLLER- AND BALL-BEARINGS. 

141. General Considerations. — By substituting rolling motion 
in bearings in place of relative sliding, friction losses can be 
greatly reduced. In the design of such bearings there are 
four points to be borne in mind: 

I. The arrangement of the parts and their form must be 
such that their relative motion is true roUing with the least 
possible amount of sliding. 

II. The form of the constraining surfaces must be such that 
the rolling parts will not have any effective tendency to leave 
their proper guides or "races." 

III. The rollers and balls must not be unduly loaded. 

IV. Provision must be made to admit the lubricant, and to 
exclude all dust and grit. 

These points will be considered in the order given. 

142. I. Rolling, Sliding, and Spinning. (See Fig. 144.) — At 




Fig. 144. 

A is shown the longitudinal section of a cylindrical ball-bearing 
of the simplest form stripped of all auxihary parts. At B 



ROLLER- AND BALL-BE A RINGS. 



211 



is shown the same for a roller-bearing. At C is a cross-section 
of either, showing but one ball or roller R. S is the journal 
and T the box. Consider T as stationary, then the point of 
contact of R and S would have the same motion relative to T 
whether considered as a point of R or of 5, and if the surface 
friction were sufficient there would be no reason for slippage. 
As a matter of fact, in the actual bearing there will be a slight 
amount of slipping .at both of the points of contact. This form 
of bearing is called the '' two-point bearing," because there are 
two points of contact. All cylindrical roller-bearings are of 
this fundamental form. In order to have them of practical use 
the rollers must be held in a case or "cage" so that their axes 
will always remain parallel with the axis of the shaft. Fig. 145 
shows such a ''cage'^ with rollers in place. 




Fig. 145. 

Since the rollers are generally of hardened and ground steel 
the best service with the least wear will be given when the 
engaging surfaces are of the same material. To meet this when 
the shaft is of soft steel, say, and the box of cast iron, a hardened 
and ground-steel ring is fitted over the shaft as a shell and 
another inside the box as a bushing, and the rollers run between 
the outer surface of the former and the inner surface of the 
latter. 

Ball-bearings are subject to an action known as "spinning." 
To illustrate this, consider the three-point ball-bearing shown 



212 



MACHINE DESIGN. 



in Fig. 146. Here the centres are as shown in B, and the conr 
ditions are correct for theoretical rolling as long as point contact 
is maintained and axis C-D remains parallel to axis E-F. But 
when the bearing is in use the points of contact, on each side 
of R, with T become small areas, as shown in B. Considering 
the relative motion of R and T at any instant it will be seen that 





Fig. 146. 



Fig. 147. 



there is an action on each side of the ball akin to that of a small 
thrust-bearing. The rubbing produced in this manner naturally 
causes undesirable friction. This is the action known as "spin- 
ning." 

Obviously it is even more marked in the case of a four- 
point bearing, as shown in Fig. 147. 

Here, also, there is pure rolling motion as long as point con- 
tact is maintained, and the axes C-D and E-F remain parallel 
to axis G-H] but as soon as the load is applied the points of 
contact become areas, and ''spinning" results at four surfaces. 
Experiments bear out the conclusion that a properly designed 
two-point bearing will have less friction than a three-point, and 
a three-point will have less than a four-point. 

In a "race " whose radius of curvature is just equal to that 
of the ball the friction becomes excessive. Such races should 
never be used. (See Fig. 148.) 

A force acting at the surface of a ball will tend to rotate 
it about an axis parallel to the tangent plane in which the actu- 



ROLLER- AND BALL-BEARINGS. 



213 



ating force lies; furthermore, this axis will be at a right angle 
with the direction of the force. This is true because it is merely 
a special application of the general law that a force applied to 
a body will tend to move it in the direction of action of the force. 







Fig. 148. 



Wl 



Fig. 149. 



The general law for the form of rolling bearings may now be 
stated as follows: 

For true rolling, the constraining surfaces of the journal 
and box {i.e., the ''races") must be so formed that the axes of 
rotation of the rollers or balls will all intersect the main axis 
of the bearing at a fixed point throughout the complete revolu- 
tion of the journal. This may be made clear by examples. 

Fig. 149 shows a ball or roller R held between two similar 
plates T and S. The upper plate, T, presses down on R with 
a force P which is transmitted through R to S. 

By the principles of so-called " rolling friction," to roll 
T on R will require a force jP (i.e., proportional to P) to 
overcome the resistance. The motion of T on i? causes R 
to roll on S, to which rolling there is induced a resistance also 
equal to jP, but in the opposite direction as regards R. These 
two forces being equal, opposite, and applied at the same dis- 
tance from the center of R, form a couple whose effect would 
be to give R a motion of rotation about an axis through its 
center, and perpendicular to the plane in which they both lie. 

This case is similar to those shown in Fig. 144, except that 



214 



MACHINE DESIGN. 



in the cases there shown S and T are not plane surfaces. Each 
ball in case A and each roller in case B tends to rotate about an 
a.xis (relatively to the '' cage," not shown) as indicated by the 
dotted lines. In both cases the individual axes all intersect 
the main axis of the journal at a fixed point, namely, at infinity, 
throughout the revolution. The general law for true rolKng 
is therefore fulfilled. 

In the cases shown in Figs. 146 and 147, obviously the 
same conditions hold. 

Next consider the thrust-bearings shown in Fig. 150: 



— --^^^^ r — ^-4^^^ r — -^\.^m\ 

^V-^-— ^ ^_ I Ic D i Q. I Ic D ! s I 



^,y^o 








Take case A first. 5 is the moving member, T the sta- 
tionary member, R one of the balls, and OF is the axis of 
rotation of S relatively to T. The center of the ball is at any 
distance r from the axis OY, and its points of contact with 5 
and T are termed A and B respectively. Relatively to the 
inclosing cage (not shown) all parts of the ball in obedience 
to the acting force tend to rotate about the axis OX which 
always cuts OF at O. It is not essential that the angle XOY 
be a right angle; it is essential that a line joining A and B 
should pass through the center of R and be perpendicular to OX, 
It is obvious that the conditions for true rolling are fulfilled. 

As OX completes one revolution about OY the ball will 
rotate about its axis A-B relative to both 5 and T. Further- 
more, since A and B become small areas under the action 
of the load, it is impossible to avoid the undesirable effect of 
*' spinning." 



ROLLER- AND BALL BEARINGS. 215 

In case -B, as S rotates relative to T, the point D common 
to R and S will have a linear velocity proportional to ^2 and, 
similarly, C's linear velocity will be proportional to ri. If 
AD and BC were two equal, independent circular disks, each 
would have true rolling motion, and BC would make ri revolu- 
tions, while AD would make ^2- But BC and AD are both 
disks of the same roller, R, and cannot rotate relative to each 
other; hence they must each make the same number of revo- 
lutions, and points C and D of the disks would have to have 
the same velocity, which is inconsistent with the conditions of 
motions of C and D as points of S. Hence a roller cannot 
be correctly used for a thrust- bearing. Short rollers securely 
held in cages are used in practice, but experiments show that 
they are not as efficient as properly designed forms * 

Consider case C. Relative to T, the double point D will 
have a linear velocity proportional to ^2 and C will have a 
linear velocity proportional to ri. Consider AD and BC as 

TiC r 

independent disks so proportioned that -ry^ =— . \i D has a 

linear velocity proportional to ^2, then the angular velocity 
of AD about its axis OX will be proportional to -jy^. 
Similarly, the angular velocity of BC about axis OX will be 



proportional to . 






Angular velocity oi AD 


r2 
7: AD 


BC r2 ri r2 


Angular velocity of BC 


ri 


AD ri r2 n 



I, 

^^ 7 1 12 fl 
TZBC 

BC n 

smce -TF, =~~- 

AD r2 

Hence the disks AD and BC have the same angular velocity 

about the axis OX, and may form parts of the same body. 

* See article by T. Hill in American Machinist, Jan. 5, 1899. Also description 
of bearing by C. R. Pratt, same periodical, June 27, 1901. 



2t6 machine design. 

This will be true of any pair of disks of the cone OBC, Any 
frustum of a cone whose apex lies anywhere on the axis OY 
will therefore fulfill the conditions for true rolling motion rela- 
tively to T when actuated by 5. 

In each of the foregoing cases the rolling members must be 
held in suitable ** cages," or they will yield to the tendency to 
displace them. 

In ball thrust-bearings it is desirable to so arrange the balls 
in the cage that each one will have a separate path, as this 
minimizes wear. 

For a three-point thrust ball-bearing the form of the races 
to permit true rolling must be as shown in Fig. 151 to be in 
accordance with the principles just demonstrated. The groove- 
angle should be as fiat as possible to reduce the friction effect 
of " spinning." 

About 120° will be found a good practicable value. 

The ball becomes akin to a cone as far as its relation with T, 
with which it has two points of contact, is concerned. It remains 
as a ball as far as its relation with 5* is concerned. In each 
case the motion imparted to the ball tends to rotate it about the 
correct axis OX and the conditions for true rolling are. sat- 
isfied. The sides of the race are tangent to the ball where it 
is cut by any line A~B which passes through O. 

A four-point ball-bearing must be designed according to 
the principles indicated in Fig. 152 for true rolling motion. 
As far as its motion reliiions with T and S are concerned, the 
ball becomes akii to a cone.* 

* The method of laying out the groove in Fig. 151 is as follows: — The axes o' 
rotation of the balls cut the main axis of the bearing at O. Draw the lower surface 
of 5 tangent to the ball at C and parallel to the ball axis OX. Draw the line OB 
cutting the ball at A and B, and draw tangent surfaces normal to the radii of the 
ball at A and B. These surfaces form the groove angle BDA. If the first trial 
gives too sharp a groove angle, increase the angle XOB and repeat the construction. 
If BDA is too flat, decrease XOB. 

For the four-point bearing shown in Fig. 152 the same method is used for deter- 
mining the groove in 5 as well as T. 



ROLLER- AND BALL-BEARINGS. 



217 



Similarly a cup and cone three-point bearing should have 
the form shown in Fig. 153. 





Fig. 153. Fig. 152. 

143. II. The form of the constraining surfaces must be such 
in ball-bearings that the balls will not have any effective tendency 
to leave their proper paths. The use of cages for this purpose 
has already been mentioned. 

If two-point bearings without cages are desired, the section 
of each race should be the arc of a circle whose radius is xV 
to J of the diameter of the ball. In two-point bearings the 
points of pressure must always be diametrically opposite. 

In three- and four-point bearings where the races are properly 
formed for true rolling, as explained in the preceding section, the 
tendency for the balls to leave the races is reduced to a minimum. 

One point, however, needs further consideration. In cup- 
and cone-bearings it is impossible to 
keep a tight adjustment at all times, 
and the least play will allow some of 
the balls on the unloaded side of the 
bearing to get out of place. 

Fig. 154 shows such a bearing 
loosely adjusted. 

The loaded cup is forced down so 
that its axis lies below the axis of 
the cone. The top ball is held correctly in place for true 





2i8 MACHINE DESIGN. 

rolling; the lower ball is free to roll to one side as seen. 
Investigation has shown that the angles a and /? should each 
be at least as great as 25° in order to return the displaced ball 
easily to its proper path by the time it becomes subjected to 
the load. If the angles are too acute there is a tendency for 
the balls to wedge in their incorrect positions, causing rapid 
wear or even crushing.* 

144. III. Allowable Loading. — Careful experiments show 
that for high efficiency and durability the loads on balls and 
rollers should be very much less than they could be with safety 
as far as their strength is concerned. f 

Let Po equal the load in pounds which is allowable for a single 
ball or roller. Then for balls 

^=diameter of ball in inches. 

K = i 500 for hardened steel balls and races, two-point bearing, 

with circular-arc races having radii equal to f J. 
K= 750 for hardened steel tails and races, three- and four- 
point bearing. 
For two-point hearing with flat races, K=<,co. 
For cast-iron lalls and races use two-fifths of these values. 
For rollers P,=Kdl. 

</ = diameter of roller in inches = mean diameter of cone, 
/ = length of roller in inches, , 
K = 40o for cast iron, 
K = joco for hardened steel. 
In thrust-bearings, if the total load=P ar.d the number of 

P 

balls = w, we have for cither balls or rollers Pj = — . 



* See article by R. Janney in American Machinist, Jan. 5, 1899. 
t See excellent article by Professor Stribeck in Z. d. V. d. I., Jan. 19 and 26, 
1901. 



ROLLER- AND BALL-BEARINGS. 21 9 

In cylindrical bearings the load is always greatest at one 
side of the bearing, the balls or rollers on the opposite side 
being entirely unloaded. It has been found that the load on 

the heaviest loaded ball or roller = Po = — P, where n is the 

n 

number of balls.* 

145. Size of Bearing. — To determine the size of the ball 
circle {i.e. the middle diameter of the 
*' race " ) given the number of balls n 
and their diameter d. (See Fig. 155.) 

r = radius of ball. i^ = radius of ball 
circle. Join the centers of two consec- 
utive balls by the chord AB = 2r. From 
the center of the ball circle, O, draw two 
radii, one to A and the other to the ^^^' ^^^' 

mid-point of A-B. Call the angle included between the radii a. 
Then 

r = i? sin a, and, 
180° 




smce a = 



R = 



n 
r 



. 180° 
sm 



* Mr. Henry Hess, of the Hess-Bright Mfg. Co., in a letter to the authors says: — 
"It is a fact, that has been determined by experience, that in radia' [i.e., cyHndrical] 
ball bearings the speed has very little influence within very wide limits. In my 
practice ... I pay no attention to speed of radial bearings up to 3000 rpm as 
influencing the load so long as such speed is fairly uniform and so long as the load 
is fairly uniform. When neither speed nor load are uniform the percussive effect 
of rapid changes must be taken in consideration; unfortunately, so far at least, the 
factors are entirely empirical and allowances are made by a comparison with 
analogous cases of previous practice. 

The case is different with thrust bearings. In these, speed is a very decided 
factor in the carrying capacity even though speed and load be uniform. Here 
again no rational formula has yet been developed to adequately represent the 
different elements, but carrying capacities for different speeds of standard bearings 



220 



MACHINE DESIGN. 



This is the radius of a circle on which the centers of the 
balls will lie when their surfaces are all in contact. It is desir- 
able to allow some clearance between the balls. This may 
be as much as 0.005 i^ch between each pair of balls provided 

the total allowance does not exceed -. When the total clear- 

4 

ance has been decided upon, it may be allowed for by making 
the actual radius of the ball circle larger than R by an amount 
one sixth of the total clearance desired. 

146. IV. Lubrication and Sealing. — On account of "spin- 
ning," faulty adjustment, and unavoidable slippage, rolling 
bearings should be properly lubricated. As they are extremely 
sensitive to the presence of dust and grit, care must be exer- 
cised that the lubricant be admitted without any danger of 
the entrance of these. 

Sealed oil-holes, dust-caps, and felt washers are commonly 
used both to retain the lubricant for bath lubrication and for 
keeping out all dirt. 

have been experimentally determined, since it was quite feasible to get different 
uniform speeds and determine under what load the carrying capacity was reached. 
We found, for instance, that for a thrust bearing employing i8 — |" balls, the 
permissible load at lo rpm was 2400 pounds; at 300 rpm — 650 pounds; at 1000 — 
450 pounds; and at 1500 only 330 pounds. We also found that, generally speaking, 
it was inadvisable to use this type of bearing for speeds materially above 1500 rpm." 
An analysis of certain standard thrust bearings in connection with the makers' 
catalog allowances for loads, gives at various speeds : — 

Load per Ball, Pounds. 



R. P.M. 


1/4" Ball. 


5/16" Ball. 


3/8" Ball. 


1500 


27.8 


35-4-44-7 


80.7 


1000 


33-3 


4I-7-57-I 


91.7 


500 


41.7 


52.1-71-4 


129 


300 


55-6 


66.7-89.2 


153 


150 


61.2 


83.3-107 


193 


10 


210 


229-339 


560 



C. R. Pratt, Trans. A.S.M.E., Vol. 27 gives as limiting load per J" ball, 100 
pounds at 700 rpm with a 6" diameter circle of rotation. 



CHAPTER XIV. 

COUPLINGS AND CLUTCHES. 

147. Couplings and Clutches Defined. — Couplings are those 
machine parts which are used to connect the ends of two shafts 
or spindles in such a manner that rotation of the one will pro- 
duce an identical rotation of the other. They are therefore 
in the nature of fastenings, and may be classified as perma- 
nent or disengaging. The latter are frequently called clutches. 

148. Permanent Couplings. — The simplest form of per- 
manent coupling is shown in Fig. 156, and is known as the 
*' sleeve " or " muff " coupling. Each shaft has a key way cut 



1 






^^^^^^^ 



fc 



-3d 

Fig. 156, 



at the end. The cast-iron sleeve of the proportions indicated is 
bored an exact fit for the shafts and has a keyway cut its entire 
length. When the sleeve is slipped over the ends of the shafts, 
the key is driven home and all relative rotation is prevented. 
The key may be proportioned according to the rules laid down 
in the chapter on Means for Preventing Relative Rotation. 



222 



MACHINE DESIGN. 



149. Flange couplings are frequently used, and Fig. 157 
illustrates the type. Approximate proportions are indicated. 



k^0.-75^i<— O.-75-H 




Fig. 157. 

The number of bolts n may be from 3 + 0.5^ to 3 + ^. Their 
diameter d^ must be such that their combined strength to resist 
a torsional moment about the axis of the shaft will be equal 
to the torsional strength of the shaft, 






16 



/a = allowable stress in outer fiber of shaft, pounds per 
square inch; 

i/ = diameter of shaft, inches; 
i? = radius of bolt circle, inches; 

/z = number of bolts ; 

c^' = diameter of bolts, inches; 

/s' = allowable shearing stress in bolts, pounds per square inch. 

This equation will approximately give 



d' = 



Vn' 



150. Compression couplings of three forms are shown in 
Figs. 158, 159, and 160. The first is similar to the ordinary 



COUPLINGS AND CLUTCHES. 



223 



sleeve coupling except that the sleeve is split in halves longi- 
tudinally and is tapered toward each end on the outside. The 
two rings shown are driven or shrunk on these tapers. 




Fig. 158. 

Instead of being held by rings the half sleeves are some- 
times bolted together as shown in Fig. 159. 




Tit Key 



Fig. 159. 

151. The " Sellers " coupling is shown in Fig. 160. An 
outer sleeve A is bored tapering from each end. A split cone 
bushing B is inserted at each end. Openings are left for three 




Section a-b 



Fig. 160. 



bolts by means of which B and B are drawn toward each other 
and thus closed down on the shaft with great force. A kev 
is also provided. 



224 



MACHINE DESIGN. 



c 



liJ 



152. Oldham's Coupling. — For all of the foregoing coup- 
lings the axes of the two shafts must be identical. Where the 

axes are parallel, Oldham's coupUng 
may be used. It is shown in Fig. 161, 
Y and consists of three parts. Each shaft 
end has keyed to it a flange or disk 
which has a diametral groove cut across 
its face. Between these two disks is 
a third which has a tongue on each face. The tongues, which 
just fit the grooves freely, run diametrically across the disk 
and are at a right angle with each other. 

153. Hooke's Coupling. — For axes which intersect, Hooke's 
coupling or " universal joint " may be used. It is shown in 
outline in Fig. 162. Each shaft has a stirrup keyed to its end. 



Fig. 161. 




Fig. 162. 

Each stirrup is connected by turning pairs to a cross-shaped 
intermediate member, the axes of whose turning pairs are 
at a right angle. For a mathematical analysis of this mechanism 
the reader is referred to Professor Kennedy's ^'Mechanics of 
Machinery." 

154. Flexible Couplings. — Where shafts which are or which 
may become slightly out of alignment are to be connected, 
some form of flexible coupling is advisable. Their principle 
is illustrated in Fig. 163. Each shaft has keyed to its end a 



COUPLINGS AND CLUTCHES. 



225 



disk which has set in its face a number of pins. The pins are 
so placed that those in the one circle will not strike those in 
the other if either shaft is rotated while the other remains at 
rest. When one shaft is to drive the other, short belts are 
placed on th£ pins as shown in B, Fig. 163. 




Fig. 163. 



155. Disengaging couplings are of two general classes: 
positive drive and friction drive. Positive-drive couplings are 
commonly called toothed or claw couplings. They consist 
of two members having projections on their faces, as shown 
in Fig. 164, which interlock when in action. A is keyed rigidly 





r 

A r— 


n — P- 


-- k 




Fig. 164. 

to its shaft. B can sHde along its shaft guided by the feather, F, 
but cannot rotate except with the shaft. When B is moved 



2 26 MACHINE DESIGN. 

in the direction of the arrow, its teeth engage with those of Aj 
and the two shafts must have the same motion of rotation. 
D shows the end of the lever which moves B. The split ring 
is bolted around B in the groove E, which it fits freely. Some- 
times short blocks which fit E are used in place of the entire 
ring. C shows an end view of the half clutch. 

Various forms of teeth may be used.* If n equals the number 
of teeth and R equals the mean radius of the clutch tooth the 
following equations may be written: 



and /«-7- = RnA j/. 



In the first equation, /s~7~, the torsional strength of the 

shaft, is equated to the crushing resistance of all the teeth 
opposed to the torsional stress. A^ is the area of the en- 
gaging face of one tooth and jc the allowable unit crushing 
stress. 

In the second equation A is the area of the root of the 
tooth subjected to shear and /«' is the allowable unit shearing 
stress. 

156. Friction couplings generally consist of two parts, a 
hollow cone, A, keyed rigidly to one shaft, and a sHding cone, 
B, held by a feather on the second shaft as seen in Fig. 165. 
By means of a lever, B can be forced against A with a con- 
siderable axial pressure. This induces friction between the 
conical surfaces, which friction resists relative rotation. To 
analyze the forces, consider Fig. 166 which shows two conical 

* See article in American Machinist, July 9, 1903. 



COUPLINGS AND CLUTCHES. 



227 



surfaces pressed together by an axial force P, The angle at the 
vertex of the cone is 2a, 




Fig. 165. Fig. 166. 

Let the coefficient of friction = /i = tan ^. 
The mean cone radius =r. 

Total pressure between the two surfaces for impending 
slippage =2R. 

Total normal pressure between two surfaces at rest =2N. 
Total friction =F=- p.2N. 

It is clear from the figure that P = 2i?sin {a-\-(j>). 
Also, N =R cos <j>. 

2N sin (a + ) 
cos 4> 



:. P 



But 2N = — , where i^=friction force between the surfaces, 

F 

.'. P= — (sin «+ M cos a). 

From this equation and the fact that the turning moment, 
M =Fr, the clutch can be designed to transmit the desired power. 

The angle a should he between 7^° and i2j° in order to avoid 
" sticking " on the one hand and too sudden seizure on the other. 

The normal pressure per sq. in. of contact area may be taken 
as 30-40 lbs. for leather or cork, and 35-45 lbs. for maple wood. 



;228 



MACHINE DESIGN. 



Table XVIII. 
fi = o.io to 0.15 for cast iron on cast iron 
= 0.15 to 0,20 for cast iron on paper 
= 0.20 to 0.30 for cast iron on leather 
= 0.20 to 0.40 for cast iron on wood. 
= 0.33 to 0.37 for cast-iron on cork. 

157. Weston Friction Coupling.— For heavy duty the prin- 
ciple of the Weston friction coupling, as shown in Fig. 167, may 




Section ELE 



Fig. 167. 



be used. The sleeve A carries two feathers on which a number 
of iron rings C can slide but not rotate. Similarly the hollow 
;sleeve B is provided with feathers which prevent the rotation 
•of the wooden rings D, while not interfering with their sliding. 




Fig. 168. 



Let there be n iron rings. Then, when B is pressed toward 
A, there will be friction induced on 2W + i surfaces. If P is 
the axial pressure and /jl the coefficient of friction, the total 



COUPLINGS AND CLUTCHES. 229 

friction F = //P(2w + i), and if r = the mean radius of the rings, 
the moment which can be transmitted =i/ =Fr. 

158. A combination friction- and claw-clutch is shown in 
Fig. 168. To start the driven shaft, B is forced to the right, 
thus bringing the friction cones into action. When the driven 
shaft has attained its proper speed, B is suddenly shifted to the 
left, thus causing the claw-clutch to engage, which gives the 
advantage of positive driving. 

Professor Bach has shown that the energy lost in friction in 
making a "running start" with a friction clutch just equals 
the kinetic energy of the shaft and attached parts at the speed 
to which they have been brought. This shows clearly the 
enormous wear and tear to which friction clutches are ordinarily 
subjected, even when designed to run practically without slip in 
the course of operation, and explains their rapid deterioration. 

For power house purposes very satisfactory magnetic clutches 
have been devised. 



CHAPTER XV. 

BELTS. 

159. Transmission of Motion by Belts. — In Fig. 169, let A 
and B be two cylindrical surfaces, free to rotate about their axes; 
let CD be their common tangent, and let it represent an inex- 
tensible connection between the two cylinders. Since it is 
inextensible, the points D and C, and hence the surfaces of the 




cylinders, must have the same linear velocity when A is rotated 

as indicated by the arrow. Two points having the same linear 

velocity, and different radii, have angular velocities which are 

inversely proportional to their radii. Hence, since the surfaces 

of the cylinders have the same linear velocity, their angular 

velocities are inversely proportional to their radii. This is true 

of all cylinders connected by inextensible connectors. Suppose 

the cylinders to become pulleys, and the tangent line to become 

a belt. Let C'Z>' be drawn; this becomes a part of the belt 

together with the portions DED^ and CFC\ making it endless, 

and rotation may be continuous. The belt will remain always 

tangent to the pulleys, and will transmit such rotation that the 

230 



BEL TS. 



angular velocity ratio will constantly be the inverse ratio of the 
radii of the pulleys. 

The case considered corresponds to a crossed belt, but the 
same reasoning applies to an open belt. (See Fig. lyo.j A and 




B are two pulleys, and CDB'C'C is an open belt. Since the 
points C and D are connected by a belt that is practically inex- 
tensible, the linear velocity of C and D is the same; therefore 
the angular velocities of the pulleys are to each other inversely 
as their radii. If the pulleys in either case were pitch cylinders 
of gears the condition of velocity would be the same. In the 
first case, however, the direction of motion is reversed, while 
in the second case it is not. Hence the first corresponds to 
gears meshing directly with each other, while the second corre- 
sponds to the case of gears connected by an idler, or to the case 
of an annular gear and pinion. While in many places positive 
driving-gears are indispensable, it is frequently the case that the 
relative position of the axes to be connected is such as would 
demand gears of inconvenient or impossible proportions, and 
belts are used with the sacrifice of positive driving. 

Of course it is necessary that a belt should have some thick- 
ness; and, since the center of pull is the center of the belt, it 
is necessary to add to the radius of the pulley half the thickness 
of the belt. The motion communicated by means of belting, 
however, does not need to be absolutely correct, and therefore 
in practice it is usually customary to neglect the thickness of the 



232 MACHINE DESIGN. 

belt. The proportioning of pulleys for the transmission of any 
required velocity ratio is now a very simple matter. 

i6o. Illustration. — A line-shaft runs 150 revolutions per min- 
ute, and is supported by hangers with 16 inches ''drop." It is 
required to transmit motion from this shaft to a dynamo to run 
1800 revolutions per minute. A 30-inch pulley is the largest 
that can be conveniently used with 16-inch hangers. Let 
X = the diameter of required pulley for the dynamo ; then from 
what has preceded x-^ 30 = 150 ^1800, and therefore ^ = 2.5 
inches. But a pulley less than 4 inches diameter should not be 
used on a dynamo.* Suppose in this case that it is 6 inches. 
It is then impossible to obtain the required velocity ratio with one 
change of speed, i.e., with one belt. Two changes of speed 




Fig. 171. 

may be obtained by the introduction of a counter-shaft. By 
this means the velocity ratio is divided into two factors. If 
it is wished to have the same change of speed from the line shaft 
to the counter as from the counter to the dynamo, then each 
velocity ratio would be \/(i8oo-m5o) =\/i2 =3.46. But this 
gives an inconvenient fraction, and the factors do not need to be 

* This limitiag size is determined mainly by considerations of thickness of 
beh required to transmit the energy, its durabihty, and its efficiency. (See § 171.) 



BELTS. 233 

equal. Let the factors be 3 and 4. (See Fig. 171.) A repre- 
sents the line-shaft, B the counter, and C the dynamo-shaft. 
The pulley on the line-shaft is 30 inches, and the speed is to be 
three times as great at the counter, therefore the pulley on the 
counter connected with the line-shaft pulley must have a diam- 
eter one third as great as that on the line-shaft = 10 inches. 
The pulley on the dynamo is 6 inches in diameter and the counter- 
shaft is to run one fourth as fast as the dynamo, and therefore 
the pulley on the counter opposite the dynamo-pulley must be 
four times as large as the dynamo-pulley = 24 inches. 

161. A belt may be shifted from one part of a pulley to 
another by means of pressure against the side which advances 
towards the pulley. Thus if, in Fig. 172, 
the rotation be as indicated by the arrow, 
and side pressure be applied at A, the belt 
will be pushed to one side, as is shown, and 
will consequently be carried into some new 
position on the pulley further to the left as it 
advances. Hence, in order that a belt may 
maintain its position on a pulley, the cen- 

FiG. 172. 
TER LINE OF THE ADVANCING SIDE OF THE 

BELT MUST BE PERPENDICULAR TO THE AXIS OF ROTATION. 

When this condition is fulfilled the belt will run and trans- 
mit the required motion, regardless of the relative position of 
the shafts. 

162. Twist Belts. — In Fig. 173 the axes AB and CD are 
parallel to each other, the above stated condition is fulfilled, 
and the belt will run correctly; but if the axis CD were turned 
into some new position, as C'D\ the side of the belt that advances 
toward the pulley E from F cannot have its center line in a 
plane perpendicular to the axis, AB, and therefore it will run off. 
But if a plane be passed through the line CD, perpendicular to 
the plane of the paper, then the axis may be swung in this plane 



234 



MACHINE DESIGN. 



in such a way that the necessary condition shall be fulfilled, and 
the belt will run properly. This gives what is known as a 
^' twist" belt, and when the angle between the shaft becomes 
90°, it is a "quarter-twist" belt. To make this clearer, see 
Fig. 174. Rotation is transmitted from ^ to J5 by an open belt, 
and it is required to turn the axis of B out of parallelism with 



.C 



/ 
/ 

/ 
/ 
/ 

/ 
/ 

// 

// 






V D' 




Fig. 173. 



Fig. 174. 



that of A. The direction of rotation is as indicated by the 
arrows. Draw the line CD. If now the line CD is supposed 
to pass through the center of the belt at C and D, it may 
become an axis, and the pulley B and the part of the belt FC 
may be turned about it, while the pulley A and the part of the 
belt ED remain stationary. During this motion the center line 
of the part of the belt CF^ which is the part that advances toward 
the pulley B when rotation occurs, is always in a plane perpen- 
dicular to the axis of the pulley B. The part ED^ since it has 
not been moved, has also its center line in a plane perpendicular 
to the axis of A. Therefore the pulley B may be swung into 
any angular position about CD as an axis, and the condition 
of proper belt transmission will not be interfered with. 

163. If the axes intersect the motion can be transmitted 



BELTS. 



235 



between them by belting only by the use of "guide'* or "idler" 
pulleys. Let AB and CD, Fig. 175, be intersecting axes, and 
let it be required to transmit motion from one to the other by 
means of a belt running on the pulleys E and F. Draw center 
lines EK and FH through the pulleys. Draw the circle, G, 
of any convenient size, tangent to the lines EK and FH. In 
the axis of the circle, G, let a shaft be placed on which are two 
pulleys, their diameters being equal to that of the circle, G. 




Fig. 175. 



These will serve as guide-pulleys for the upper and lower sides 
of the belt, and by means of them the center lines of the advanc- 
ing parts of both sides of the belt will be kept in planes perpen- 
dicular to the axis of the pulley tow^ard which they are advancing, 
the belts will run properly, and the motion will be transmitted, 
as required. 

The analogy between gearing and belting for the trans- 
mission of rotary motion has been mentioned in an earlier 
paragraph. Spur-gearing corresponds to an open or crossed 
belt transmitting motion between parallel shafts. Bevel-gears 
correspond to a belt running on guide-pulleys transmitting 
motion between intersecting shafts. Skew bevel and spiral gears 
correspond to a "twist" belt transmitting motion between shafts 
that are neither parallel nor intersecting. 



E/F; 



236 MACHINE DESIGN. 

164. Crowning Pulleys. — If a flat belt is put on a "crowning " 
^ pulley, as in Fig. 176, the tension on AB will be 

greater than on CD. The belt lying flat Sit AC will 

— n have its approaching portion bent out as indicated by 

AE and CF, and as rotation goes on the belt will 

be carried toward the high part of the pulley, i.e., 

it will tend to run in the middle of the pulley. 

Fig ^176 This is the reason why nearly all belt pulleys, except 

those on which the belt has to be shifted into different 

positions, are turned ''crowning." 

165. Cone Pulleys. — In performing different operations on a 
machine or the same operations on materials of different degrees 
of hardness, different speeds are required. The simplest way 
of obtaining them is by use of cone pulleys. One pulley has a 
series of steps, and the opposing pulley has a corresponding 
series of steps. By shifting the belt from one pair to another 
the velocity ratio is changed. Since the same belt is used on 
all the pairs of steps, these must be so proportioned that the 
belt length for all the pairs shall be the same ; otherwise the belt 
would be too tight on some of the steps and too loose on others. 
Let the case of a crossed belt be first considered. The length 
of a crossed belt may be expressed by the following formula: 
Let Z= length of the belt; d= distance between centers of rota- 
tion; i^= radius of the larger pulley; r=radius of the smaller 
pulley. (See Fig. 177.) Then 



L = 2 V ^2 _ (i^ + rf + (i? + r) (tt + 2 arc whose sine is (R + r)-^d), 

In the case of a crossed belt, if the size of steps is changed so 
that the sum of their radii remains constant, the belt length 
will be constant. For in the formula the only variables are R 
and r, and these terms only appear in the formula SiS R + r; but 
R + r is by hypothesis constant. Therefore any change that is 



BELTS. 



23 



made in the variables R and r, so long as their sum is constant, 
will not affect the value of the equation, and hence the belt 
length will be constant. It will now be easy to design cone 
pulleys for a crossed belt. Suppose a pair of steps given to 
transmit a certain velocity ratio. It is required to find a pair 
of steps that will transmit some other velocity ratio, the length 
of belt being the same in both cases. Let R and r = radii of the 
given steps; R' and /= radii of the required steps; R-{-r = 
R' +f' =a\ the velocity ratio of R to r' =h. There are two 




Fig. 177. 

equations between R and r', R' -^r^ =h, and R^ + /=a. Com- 
bining and solving, it is found that r' =a-=-(i +&), and R' =a — r\ 
For an open belt the formula for length, using same symbols 
as for crossed belt, is 



L = 2V(P-{R-rf + 7z{R-\-r) 

■\-2{R — r) (arc whose sine is {R — r)^d), 

If R and r are changed as before {i.e., R + r = R^-\-/=2i 
constant), the term R — r w^ould of course not be constant, and 
two of the terms of the equation would vary in value; therefore 
the length of the belt would vary. The determination of cone 
steps for open belts therefore becomes a more difficult matter, 
and approximate methods are almost invariably used. 

166. Graphical Method for Cone-pulley Design. — The fol- 
lowing graphical approximate method is due to Mr. C. A. Smith, 
and is given, with full discussion of the subject, in ''Transactions 



238 



MACHINE DESIGN. 



of the American Society of Mechanical Engineers," Vol. X, 
p. 269. Suppose first that the diameters of a pair of cone steps 
that transmit a certain velocity ratio are given, and that the 
diameters of another pair that shall serve to transmit some 
other velocity ratio are required. The distance between centers 
of axes is given. (See Fig. 178.) Locate the pulley centers O 




Fig. 178. 

and O' at the given distance apart; about these centers draw 
circles whose diameters equal the diameters of the given pair of 
steps; draw a straight line GH tangent to these circles; at 7, 
the middle point of the line of centers, erect a perpendicular, 
and lay off a distance JK equal to the distance between centers, 
C, multiplied by the experimentally determined constant 0.314; 
about the point K so determined, draw a circular arc AB tan- 
gent to the line GH. Any line drawn tangent to this arc will be 
the common tangent to a pair of cone steps giving the same 
belt length as that of the given pair. For example, suppose 
that OD is the radius of one step of the required pair; about O, 
with a radius equal to OD, draw a circle ; tangent to this circle 
and the arc AB draw a straight line DE; about O' and tangent 
to DE draw a circle } its diameter will equal that of the required 
step. 

But suppose that) instead of having one step of the required 
pair given, to find the other corresponding as above, a pair of 



BELTS. 239 

steps are required that shall transmit a certain velocity ratio, 
= r, with the same length of belt as the given pair. Suppose 
OD and O'E to represent the unknown steps. The given velocity 

ratio equals r. Also, r=jrp^. But from similar triangles 

FO 
OD^O'E=FO^FO\ Therefore r=y^; but FO-C + x, 

C + x C 

and FO' =x. Therefore r= , and x= . Hence with 

X r—i 

r and C given, the distance x may be found, and a point F located, 

such that if from F a line be drawn tangent to AB, the cone 

steps drawn tangent to it will give the velocity ratio, r, and a 

belt length equal to that of any pair of cones determined by a 

tangent to AB. The point F often falls at an inconvenient 

distance. The radii of the required steps may then be found 

as follows: Place a straight-edge tangent to the arc AB and 

measure the perpendicular distances from it to O and 0\ 

The straight-edge may be shifted until these distances bear 

the required relation to each other. 

167. Design of Belts. — Fig. 179 represents two pulleys 

connected by a belt. When no moment is applied tending to 

produce rotation this tension in the two sides of the belt is 

practically equal. Let T^ represent this tension. If now 

an increasing moment, represented by Rl, be applied to the 

driver, its effect is to increase the tension in the lower side of 

the belt and to decrease the tension in the upper side. With 

the increase of Rl this difference of tension increases till it is 

equal to P, the force with which rotation is resisted at the surface 

of the pulley. Then rotation begins* and continues as long as 

this equaHty continues; i.e., as long as Ti — To^P, in which 

ri= tension in the driving side and r2= tension in the slack 

* While the moving parts are being brought up to speed the difference of 
tension must equal P plus force necessary to produce the acceleration. 



240 



MACHINE DESIGN. 



side. (See Fig. i8o.) The tension in the driving side is increased 



at the expense of that in the slack side. Therefore — 



Ti + T, 



To find the value of 7^. The increase in tension from the 
^ 2 

slack side to the driving side is possible because of the frictional 

resistance between the belt and pulley surface. Consider any 

Fig. 179. 




Fig. 180. 

element of the belt, ds, Fig. 181. It is in equilibrium under 
the action of the following forces: 

T, the value of the varying tension at one end of ds; 

T + dT, the value of the varying tension at the other end of ds: 

cds, the centrifugal force ; 

pds, the pressure between the face of the pulley and ds; 

dF = fipds, the friction between the element of belt and 
pulley face. 

Let the value of cds be first determined. 

Let A = sectional area of belt in square inches; 
r = total tension at any section in pounds; 



BELTS, 



24 




t= allowable working tension in pounds per square inch 

T 

of cross-section of belt, so that A= — \ 

I 

w = weight of belt per cubic inch; 

V = velocity of belt in feet per minute ; 

Z' = velocity of belt in feet per second; 
i^ = radius of pulley in feet; 

r = radius of pulley in inches; 

g = acceleration due to gravity = 32.2 feet per second per 

second ; 
C = centrifugal force per cubic inch of belt in pounds; 

c = centrifugal force per linear inch of belt in pounds. 



'42 ' MACHINE DESIGN, 

^ .^ , . mass X velocity^ 

Centrifugal force = —7: 7 : — • 

° radius of curvature 

w iP- 
C = —-^ in foot-pounds-second system. 

w v'^ T w v^ \2wv^ T 



:, c = A 
As an abbreviation let z = 



g R t g R gt r 



gt 



z z 

:.c = T—, and cds = T—ds. 
r r 

T 

Now, to find the value of 77^: 

i 2 

Since dT is small compared with T, no appreciable error is 

introduced by considering pds as acting normally to the chord 

of ds, and we may write the equation (sum of vert. comp. =0, 

in Fig. 181) 

. dd J dd\ 

pds + cds = r sm — + (T + c^T) I sin — 1 . 

a = total arc of contact in degrees. 

6 is the total arc of contact in radians =0.01750:. 

. dd . dd 

dd is so small that sin — may be considered equal to — in 

radians. Also dT and dd both being very small compared to 
the other quantities, any terms containing their product may be 
dropped. Therefore 

pds + cds = Tdd. 
:. pds = Tdd - cds = Tdd - T-ds. 



But ds^rdd. 



.-. pds = Tdd - Tzdd = T{i-z) dd. 



BELTS. 243 

Summing the moments about o (Fig. 181) gives 
T^-dT = T + npds. 
,'. dT = fipds = /jLT(i—z)dd. 



I 



dT 
T 

^^dT 
T 



Y = Ki-z)dd, 



= ;i(i-z)£dd. 



T 

^ 2 

T 
and common log -7^ = . 4242/1(1— z)d. 

^ 2 

The following equations are now established: 

Ti-T2=P, (i) 

Ti + T2 = 2Ts, (2) 

T 

log~ = .4^4Sfi{i-z)d (3) 

J^ 2 

When the belt speed is under 2000 feet per minute the value 
of z is so small that it may be neglected, and equation (3) becomes 

"T 

log ~ = •4343/^<^- 
i 2 

The right-hand members of (i) and (3) can usually be deter- 
mined; hence the value of T (the maximum stress in the belt) 
may be found, and proper proportions may be given to the belt. 
If W foot-pounds per minute are to be transmitted, and the 
velocity of the rim of the pulley transmitting this power in feet 
per minute equals V, then the force P equals the work divided 

W 

by the velocity, or P = ^. The value of a is found from the 

diameters of the pulleys and their distance between centers, 



244 



MACHINE DESIGN. 



and may usually be estimated accurately enough. The value 
of /x, the coefficient of friction, varies with the kind of belting, 
the material and character of surface of the pulley, and with 
the rate of slip of the belt on the pulley. Experiments made 
at the laboratory of the Massachusetts Institute of Technology, 
under the direction of Professor Lanza, indicate that for leather 
belting running on turned cast-iron pulleys, the rate of slip for 
efficient driving is about 3 to 4 feet per minute; and also that 
the coefficient of friction corresponding to this rate of slip is 
about 0.27. The value 0.3 may be used. If this value of jj. 
be used, the slip will be kept within the above limits, if the belt 

be put on with a proper initial tension, = 7^3 = , and the 

driving of the belt so designed will be satisfactory. 

The weight of leather belting on the average per cubic inch 
= w = 0.035 lb. 

The value of the allowable tension varies with the quahty of 
the belt, the nature of the spHce, and other items, t = 300 lbs. 
per square inch is a good safe value for ordinary conditions. 
This is about equivalent to 70 lbs. per inch of width of single 
belt. The following table of values of z has been computed 
with these values of w and /. V is the velocity of the belt in 
feet per minute, and v is the velocity in feet per second. 

Table XIX. — Values of 2. 



V ... 


1800 


2400 


3000 


3600 


4200 


4800 


5400 


6000 


6600 


7200 


7800 


V. . . . 


30 


40 


50 


60 


70 


80 


90 


100 


no 


120 


130 


z. . . . 


•039 


.070 


.118 


•157 


.214 


.279 


•352 


•435 


.526 


.626 


•735 



It becomes evident that i — z decreases rapidly as the velocity 
of the belt increases. In fact, unless a greater value of / than 
300 lbs. per square inch is used a leather belt weighing 0.035 lb. 
per cubic inch will cause z to have a value of unity at about 



BELTS. 245 

9000 feet per minute. This would make the term i—z reduce 
to zero. In other words the belt can transmit no power at this 
speed because the centrifugal force is so great that no pressure 
exists between the belt and the face of the pulley and hence 
there is no friction.* 

168. Problem. — A single-acting pump has a plunger 8 inches 
= 0.667 foot in diameter, whose stroke has a constant length of 
10 inches =0.833 foot. The number of strokes per minute is 50. 
The plunger is actuated by a crank, and the crank-shaft is 
connected by spur-gears to a pulley-shaft, the ratio of gears 
being such that the pulley-shaft runs 300 revolutions per minute. 
The pulley which receives the power from the line- shaft is 18 
inches in diameter. The pressure in the delivery-pipe is 100 lbs. 
per square inch. The line-shaft runs 150 revolutions per 
minute, and its axis is at a distance of 12 feet from the axis of 
the pulley-shaft. 

Since the line-shaft runs half as fast as the pulley-shaft, the 

diameter of the pulley on the line-shaft must be twice as great 

as that on the pulley-shaft, or 36 inches. The work to be done 

per minute, neglecting the friction in the machine, is equal to 

the number of pounds of water pumped per minute multiplied 

by the head in feet against which it is pumped. The number 

of cubic feet of water per minute, neglecting ''slip," equals the 

displacement of the plunger in cubic feet multipKed by the 

0.6672x7: 

number of strokes per mmute = Xo.833X5o = i4.55, 

4 

and therefore the number of pounds of water pumped per minute 
= 14.55X62.4 = 908. One foot vertical height or "head" of 
water corresponds to a pressure of 0.433 ^b. per square inch, 
and therefore 100 lbs. per square inch corresponds to a "head" 

* The method of computing belts in this seciion is open to minor theoretical 
objections. Its value lies in the fact that it gives equations of a simple form 
which can be applied easily and which give results well within the limits of 
accuracy set by the variations ki the physical preperties of the belt materials. 



246 



MACHINE DESIGN. 



of 100-^0.433=231 feet. The work done per minute in pump- 
ing the water therefore is equal to 908 lbs. X 231 feet =209,748 
ft.-lbs. The velocity of the rim of the belt-pulley is equal to 
300 XI. 5 X7r= 1414 feet per minute.* Therefore the force P = Ti 
— 7*2 = 209,748 ft.-lbs. per minute -M414 feet per minute = 
148 lbs. 



-s 




§ 1 


^ 


II 

q: 


T 


V f- 


^ 




r 


) ^ 


^ 



Fig. 182. 



R-r o" 

To find a (see Fig. 182) sin/9 = — r— = 77 = 0.0625. 

Therefore /? = 3°35'; a = i8o°-2/3 = i8o°-7° 10^ = 172° 50'; a 

in 71 measure = 172! X0.0175 =3.025 =6. 
T 
log Y^ = 0.4343 X/^X^ =0.4343X0.3X3.02 5 =0.3941. 

/. ^ = 2.48; P = ri-r2 = i48. 

Combining these equations Ti is found to be equal to 248 lbs., 
the maximum stress in the belt. 

The cross-sectional area of belt should be equal to — = 

t 300 

0.83 square inch. 

Single-thickness belting varies from 0.2 to 0.25 of an inch in 

thickness, hence the width called for by our problem would be 

0-83 
0.2 



4 inches, say. 



169. Problem. — A sixty -horse-power dynamo is to run 1500 
revolutions per minute and has a 15-inch pulley on its shaft. 



At this speed the simple form of the belt formula may be used. 



BELTS. 247 

Power is supplied by a line-shaft running 150 revolutions per 
minute. A suitable belt connection is to be designed. 

The ratio of angular velocities of dynamo-shaft to line-shaft 
is 10 to I ; hence the diameter of the pulley on the line-shaft 
would have to be ten times as great as that of the one on the 
dynamo, =12.5 feet, if the connections were direct. This is 
inadmissible, and therefore the increase in speed must be ob- 
tained by means of an intermediate or counter shajL Suppose 
that the diameter of the largest pulley that can be used on the 
counter-shaft =48 inches. Then the necessary speed of the 

counter-shaft = i5ooX-^ = 47o, nearly. The ratio of diameters 

of the required pulleys for connecting the line-shaft and the 

470 
counter-shaft = = 3-i3- Suppose that a 60-inch pulley can 

be used on the line-shaft, then the diameter of the required 

pulley for the counter-shaft will = =19 inches, nearly. Con- 

sider first the belt to connect the dynamo to the counter- shaft. 
The work =60X33,000 = 1,980,000 ft. -lbs. per minute; the rim 

of the dynamo-pulley moves X 1500 = 5890 feet per minute. 

Therefore Ti — T2= — 5 = 336 lbs. The axis of the counter- 

5890 

shaft is 10 feet from the axis of the dynamo, and, as before, 

. „ R — r 24 — 7. s 
sm/9 = — j- = — — ^-^=0.1375. 
^ / 120 ^'^ 

Therefore /? = 7° 54'. 

a = 180° -2/? = 164° 12', 

<9 = i64°.2Xo.oi75 =2.874. 

The nearest value of V, in the table for values of z (p. 244), 
to 5890 is 6000, and the corresponding value of z is 0.435. 
Hence 1—2 = 0.565. 

log ^ = 0.4343X0.3X0.565X2.874 =0.2111; 



248 MACHINE DESIGN. 

J^ 2 

Ti =i.626r2. 

But ri-r2 = 336 lbs. 

/. o.626r2 = 336 lbs. 

.-. 2^2 = 537 lbs. 

.-. ri=873lbs. 

873 

The cross-sectional area of the belt = =2.0 square inches. 

300 ^ ^ 

A double belt is about | inch thick. Our problem, then, 

calls for a double belt 1X2.9=8 inches, say, wide. 

170. Variation of Driving Capacity. — From equation (3), 

T 

p. 231, it follows that the ratio of tensions, — , when the belt 

^ 2 

slips at a certain allowable rate {i.e., when fi is constant), de- 
pends only upon a. The velocity of the belt also remains 
constant. This ratio, therefore, is independent of the initial 

T 
tension, Tz\ hence ''taking up " a belt does not change — . 

T2 

The difference of tension, Ti — T2=P, is, however, dependent 

on Ts. Because p, the normal pressure between belt and 

pulley, varies directly as T3, then, since dF ^ /jLpds = dT, it 

follows that dT varies with T3, and hence 

fdT = Ti-T2=P 
varies with Ts. This is equivalent to saying that ''taking up " 
a belt increases its driving capacity. 

This result is modified because another variable enters the 
problem. If Ts is changed, the amount of slipping changes, 
and the coefficient of friction varies directly with the amount 
of slipping. Therefore an increase of Ts would increase p 



BELTS. 249 

and decrease 11 in the expression upds=dT, and the converse 
is also true. This is probably of no practical importance. 

The value of P may also be increased by increasing either /£, 
the coefficient of friction, or 6, the arc of contact, since increase 

of either increases the ratio 7^, and therefore increases 

J^ 2 

Ti-T2=P. 

Increasing Ts decreases the life of the belt. It also in- 
creases the pressure on the bearings in which the pulley-shaft 
runs, and therefore increases frictional resistance; hence a 
greater amount of the energy supplied is converted into heat and 
lost to any useful purpose. But if Ts is kept constant, and /i 
or 6 is increased, the driving power is increased without increase 
of pressure in the bearings, because this pressure = 2 T^a remains 
constant. When possible, therefore, it is preferable to increase 
P by increase of // or d, rather than by increase of T^. 

Application of belt-dressing may serve sometimes to in- 
crease /x. 

If, as in Fig, 179, the arrangement is such that the upper 
side of the belt is the slack side, the ''sag " of the belt tends to 

T 

increase the arc of contact, and therefore to increase 7^. If 

the lower side is the slack side, the belt sags away from the pulleys 
and 6 and -^ are decreased. 

-L 2 




Fig. 183. 

An idler-pulley, C, may be used, as in Fig. 183. It is pressed 

against the belt by some means. Its purpose may be to increase 

T^ -f T2 
P by increasing the tension, Ts = . In this case friction 



250 MACHINE DESIGN. 

in the bearings is increased, and this method should be avoided. 

Or it may be used on a slack belt to increase the angle of con- 

T 
tact, a, the ratio 7^, and therefore P, the driving force. In 
-L 2 

7^ 4- 7^ 
this case the value of T3, =— — , may be made as small a 

value as is consistent with driving, and hence the journal friction 
may be small. 

Tighteners are sometimes used with slack belts for dis- 
engaging gear, the driving-pulley being vertically below the 
follower. 

In the use of any device to increase pt and a, it should be 
remembered that Ti is thereby increased, and may become 
greater than the value for which the belt was designed. This 
may result in injury to the belt. 

In Fig. 184, the smaller pulley. A, is above the larger one, 
B. A has a smaller arc of contact, and hence the belt would 




Fig. 184. Fig. 185. 

slip upon it sooner than upon B. The weight of the belt, how- 
ever, tends to increase the pressure between the belt and A, 
and to decrease the pressure between the belt and B. The 
driving capacity of A is thereby increased, while that of B is 
diminished; or, in other words, the weight of the belt tends to 
equalize the inequality of driving power. If the larger pulley 
had been above, there would have been a tendency for the 
belt weight to increase the inequality of driving capacity of the 



BELTS. 251 

pulleys. The conclusion from this, as to arrangement of pulleys, 
is obvious. 

171. Proper Size of Pulleys. — A belt resists a force which 
tends to bend it. Work must be done, therefore, in bending a 
belt around a pulley. The more it is bent the more work is 
required and the more rapidly the belt is worn out. Suppose 
AB^ Fig. 185, to represent a belt which moves from A toward 
B. If it runs upon C it must be bent more than if it runs 
upon D. The work done in bending the belt is converted into 
useless heat by the friction between the belt fibers. It is desirable, 
therefore, to do as little bending as possible. This is one reason 
why large pulleys in general are more efficient than little ones. 
The resistance to bending increases with the thickness of the 
belt, and hence double belts should not be used on small pulleys 
if it can be avoided. 

Double belts may be used on pulleys 12" and over. 
Triple " " " '' '' " 20'' " 

Quadruple '' " '' '' '' '' 30'' '' 

172. Distance Desirable between Shafts. — In the design of 
belting care should be taken not to make the distance between 
the shafts carrying the pulleys too small, especially if there is the 
possibility of sudden changes of load. Belts have some elasticity, 
and the total yielding under any given stress is proportional to 
the length, the area of cross-section being the same. There- 
fore a long belt becomes a yielding part, or spring, and its yielding 
may reduce the stress due to a suddenly applied load to a safe 
value ; whereas in the case of a short belt, with other conditions 
exactly the same, the stress due to much less yielding might be 
sufficient to rupture or weaken the joint. 

173. Rope-drives. — The formulae which have been derived 
for belts also apply to rope-drives. For good durabihty the 
allowable tension in a rope-drive should be about 200^/2 \\^^^ 
where d is the diameter of the rope in inches. Experiments 



252 



MACHINE DESIGN. 



vary greatly in the value of the coefficient of friction for a well- 
lubricated rope on a fiat-surfaced smooth metal pulley. It may 
be taken equal to 0.12.* But ropes are not commonly used 
on flat pulleys; instead of this they run in grooves on the faces 
of sheave wheels, and substitution must be made for /z in the 

r 1 / \ , angle of groove 
formula (3), not 0.12 but o.i2Xcosec . 

The following table gives the values of jn for different angles 

of grooves. 

Table XX. 



Angle of groove in degrees . . 
/" 



30 


35 


40 


45 


50 


55 


0.46 


0.40 


0.35 


0.31 


0.28 


0.26 



60 

0.24 



Taking the weight of rope per linear inch =o.o26Sd^ lbs., and 
the allowable tension =200^^ lbs., and solving for z at various 
speeds, gives the following results: 

Table XXI. 



V... 
i—z. 



1000 
0.08 



2000 
0.94 



2500 
0.91 



3000 
0.87 



3500 
0.83 



4000 
0.78 



4500 
o. 72 



0.65 



5500 
0.58 



6000 

0.50 



6500 

0.41 



7000 

0.32 



7500 

0.22 



8000 

O. II 



8500 

0.00 



V is the velocity of the rope in feet per minute. 
For convenience the following table is given showing the 
corresponding values of angles in degrees and circular measure: 



Table XXII. 



105 

1-83 1 

1 


120 

2.09 


135 

2-35 


150 
2.62 


165 
2.88 


180 
3-14 


195 
3-43 


210 
3.66 



240 

4.19 



It will be remembered that ^=0.01750:. 

The diameter of the sheave wheel is properly calculated from 
the point of tangency of the rope to the groove {A- A, Fig. 186) 
and not from the middle of the rope O. The diameter of the 
sheave wheel should not be too small or the rope will wear out 



Rope -driving," by J. J. Flather. New York: Wiley & Sons. 



BELTS. 



253 



very rapidly. The following table gives the minimum values, 
D being the diameter of the wheel and d that of the rope : 

Table XXIII. 



24 


36 


48 


60 


If- 
72 



It is still better to make D^^od. 

Two systems of rope-driving are in use, the 
EngHsh and the American. In the former a number 
of ropes are used side by side. In the latter a 
single continuous rope is used with a guide and 
tightener. As long as all the grooves in each 
sheave are alike, each rope will tend to carry its 
proportionate share of the load in the EngHsh 
system, provided all the ropes had the same 
original tension, and have stretched the same 
amount. 

For the American system the grooves in the 
larger sheave should have a greater angle than 
those in the smaller sheave or the load will be 
unequally divided among the various wraps of the 
rope.* 

The tension in each wrap of the rope will be the 
same, when running, if the friction on each sheave 
wheel is the same. The friction on each pulley 
will be the same if the products of the arcs 
of contact by the respective coefficients are 
equal. 

r. 



Fig. 



Let // = coefficient of larger sheave =0.12 cosec 

7-= angle of groove of larger sheave; 
a:=arc of contact of larger sheave; 

* See further Proc. Am. Soc C. E., Vol. XXIII. Mr. Spencer Miller on 
Rope-driving. 



254 



MACHINE DESIGN. 



ju' = coefficient of smaller sheave =0.12 cosec — ; 

f = angle of groove of smaller sheave; 
a' = arc of contact of smaller sheave. 
Then jia should =ii!a! . 

r r' ol' 

.'. cosec — = cosec — x— . 
2 2 0: 

The following table gives the proper values for equal ad- 
hesion : 

Table XXIV. — ^Angle of Groove for Equal Adhesion. 



Arc of contact on small pulley a' 
Arc of contact on large pulley a ' ' ' 
Angle of groove in large pulley when 

groove in small pulley = 35° 

Angle of groove in large pulley when 

groove in small pulley = 40° 

Angle of groove in large pulley when 

groove in small pulley = 45° 



0.9 


0.8 


0.75 


0.7 


0.65 


40° 


44° 


47° 


51° 


55° 


45° 


50° 


54° 


58° 


64° 


50° 


55° 


60° 


66° 


72° 



60^ 

70^ 

80= 



The angle of groove on the smaller sheave wheel is generally 
made 45°. Assuming this an angle of contact of 165°, and an 
allowable stress = 2ood^, the following table has been com- 
puted for the horse -power transmitted by each wrap of the rope: 



Table XXV. 



Velocity 
of Rope 
in Feet 






Diameter of Rope 


in Inches. 




















per 
















Minute, 
V. 


i 


i 


I 


il 


li 


li 


2 


1000 


1.24 


2 .25 


3-57 


5-59 


8.02 


10.85 


14.20 


2000 


2.70 


3-84 


6 


84 


10.68 


15-39 


20.93 


27.36 


2500 


3 -30 


4.71 


8 


3« 


13.10 


18.86 


25.66 


33-54 


3000 


3-^3 


5-46 


9 


80 


15-39 


21.87 


29.74 


38.88 


3500 


4-30 


6.23 


II 


09 


^7-33 


24.94 


34.03 


44 ■ 35 


4000 


4-74 


6.83 


12 


15 


18.98 


27-33 


37-17 


48.59 


4500 


5.01 


7.24 


12 


89 


20.15 


29.00 


39-45 


5^-57 


5000 


5.20 


7-47 


13 


29 


20.76 


29.89 


40 . 65 


53 ^5 


5500 


5-29 


7.60 


13 


53 


21.14 


30-43 


41-39 


54-11 


6000 


5.C8 


7-32 


13 


10 


20.36 


29.32 


39-77 


52.12 


6500 


4-74 


6.83 


12 


13 


19.00 


27-34 


37.21 


48.63 


7000 


4.12 


5-93 


10 


54 


16.47 


23.72 


32.26 


42.18 


7500 


3-25 


4.67 


8.32 


13.00 


18.72 


25.42 


33-23 



BELTS. 255 

For durability a few turns of a larger rope are preferable to 
more turns of a smaller rope. 

The most economical speed, taking first cost and relative 
wear into consideration, is about 4500 feet per minute. 



CHAPTER XVI. 

FLY-WHEELS. 

174. Theory of Fly-wheel. — Often in machines there is 
capacity for uniform effort, but the resistance fluctuates. In 
other cases a fluctuating effort is applied to overcome a uniform 
resistance, and yet in both cases a more or less uniform rate 
of motion must be maintained. When this occurs, as has been 
explained,* a moving body of considerable weight is interposed 
between effort and resistance, which, because of its weight, 
absorbs and stores up energy with increase of velocity when 
the effort is in excess, and gives it out with decrease of velocity 
when the resistance is in excess. This moving body is usually 
a rotating body called a fly-wheel. 

To fulfill its office a fly-wheel must have a variation of 
velocity, because it is by reason of this variation that it is able 
to store and give out energy. The kinetic energy, E, of a 
body whose weight is W, moving with a velocity i;, is expressed 
by the equation 

iL = . 

To change E, with W constant, v must vary. The allowable 
variation of velocity depends upon the work to be accomplished. 
Thus the variation in an engine running electric lights or spin- 
ning-machinery should be very small, probably not greater 

* See § 43. 

256 



FLY-IVHEELS. 257 

than a half of one per cent, while a pump or a punching- 
machine may have a much greater variation without interfering 
with the desired result. If the maximum velocity, v\^ of the 
fly-wheel rim and the allowable variation are known, the mini- 
mum velocity, 1^2, becomes known; and the energy that can be 
stored and given out with the allowable change of velocity L« 
equal to the difference of kinetic energy at the two velocities. 

E = ^ — (vi^—V2V' 

2g 2g 2g 

175. The general method for fly- wheel design is as follows; 
Find the maximum energy due to excess or deficiency of effort 
during a cycle of action, =E. Use the foot-pound-second 
system of units. Assume a convenient mean diameter of fly- 
wheel rim. From this and the given maximum rotative speed 
of the fly-wheel shaft find Vi. Solve the above equation for 
IF thus: 

2gE 



W = 



V^—V'^' 



Substitute the values of E, V\, v^, and ^ = 32.2 feet p^r second,^ 
whence W becomes known, = weight of fly-wheel rim. The 
weight of rim only will be considered; the other parts of the 
wheel, being nearer the axis, have less velocity and less capacity 
per pound for storing energy. Their effect is to reduce slightly 
the allowable variation of velocity.* 

176. Problem. — In a punching-machine the belt is capable 
of applying a uniform torsional effort to the shaft; but most 

* Numerical examples taken from ordinary medium-sized steam-engine fly- 
wheels show that while the combined weight of arms and hub equals about one 
third of the total weight of the wheel, che energy stored in them for a given varia- 
tion of velocity is only about 10 per cent of that stored in the rim for the same 
variation. 



25S MACHINE DESIGN. 

of the time it is only required to drive the moving parts of the 
machine against frictional resistance. At intervals, however, 
the punch must be forced through metal which offers shearing 
resistance to its action. Either the belt or fly-wheel, or the two 
combined, must be capable of overcoming this resistance. A 
punch makes 30 strokes per minute, and enters the die J inch. It 
is required to punch |-inch holes in steel plates J inch thick. 
The shearing strength of the steel is about 50,000 lbs. per square 
LQch. When the punch just touches the plate the surface which 
offers shearing resistance to its action equals the surface of the 
hole which results from the punching, =7idt, in which J = diam- 
eter of hole or punch, / = thickness of plate. The maximum 
shearing resistance, therefore, equals 711x^X50,000 = 58,900 lbs. 
As the punch advances through the plate the resistance decreases 
because the surface in shear decreases, and when the punch 
just passes through the plate the resistance becomes zero. If 
the change of resistance be assumed uniform (which would 
probably be approximately true) the mean resistance to punching 
would equal the maximum resistance + minimum resistance, ^2, 

= =29,450. The radius of the crank which actuates 

the punch = 2 in. In Fig. 187 the circle represents the path of 
the crank-pin center. Its vertical diameter then represents the 
travel of the punch. If the actuating mechanism be a slotted 
cross-head, as is usual, it is a case of harmonic motion, and it may 
be assumed that while the punch travels vertically from A to B, 
the crank-pin center travels in the semicircle ACB. Let BD 
and DE each = J inch. Then when the punch reaches E it 
just touches the plate to be punched, which is J inch thick, and 
when it reaches D it has just passed through the plate. Draw the 
horizontal lines EF and DG and the radial lines OG and OF. 
Then, while the punch passes through the plate, the crank-pin 
center moves from F to G, or through an angle (in this case) 



FLY -IVH EELS. 



259 



of 19°. Therefore the crank-shaft A^ Fig. 188, and attached 

gear rotate through 19° during the action of the punch. The 

ratio of angular velocity of the pinion and the gear = the inverse 

60 
ratio of pitch diameters =— =5. Hence the shaft B rotates 



12 



through an angle =19^X5 =95° during the action of the punch. 
If there were no fly-wheel the belt would need to be designed 
to overcome the maximum resistance; i.e.^ the resistance at the 
instant when the punch is just beginning to act. This would 



Tt 




Crank 



m 



.1- 



i 

•■a 

di 



Fig. 188. 



give for this case a double belt about 20 inches wide. The need 
for a fly-wheel is therefore apparent. Assume that the fly- 
wheel may be conveniently 36 inches mean diameter, and that 
a single belt 5 inches wide is to be used. The allowable maxi- 
mum tension is then =5Xallowable tension per inch of width 
of single belting = 5X70 = 350 lbs. = ri. 1 

Since the pulley-shaft makes 150 revolutions per minute and 
the diameter of the pulley is 2 feet, the velocity of the belt = 
150X2X71 = 942 feet per minute. At this slow speed the simple 

7^ 
form of the belt formula may be used, i.e., log Tfr =0.4^,47,110. 

-t 2 

Assume an angle of contact of 180°. Then 



26o MACHINE DESIGN. 

^=3.1416, 

log — =0.4081; 

-t 2 

and ,.1=2.56. 

-^ 2 

^^=5-36.7 lbs. 

Ti — T2 = 213.3 lbs. =the driving force at the surface of the pulley. 
Assume that the frictional resistance of the machine is 
equivalent to 25 lbs. applied at the pulley-rim. Then the belt 
can exert 213.3 — 25=188.3 lbs. =P, to accelerate the fly-wheel 
or to do the work of punching. Assume variation of velocity = 10 
percent. The work of punching = the mean resistance offered 
to the punch multipHed by the space through which the punch 

acts, = X 0.5" = 14725 in. -lbs. =1225 ft. -lbs. The pulley- 

shaft moves during the punching through 95°, and the driv- 
ing tension of the belt, =P = 188.3 ^^s., does work=PXspace 

moved through during the punching = 188.3 lbs. X;rG?--7- = 188.3 

lbs.X7rX2 ft.X-^ = 3i2 ft.-lbs. The work left for the fly- 
360 ^ ^ 

wheel to give out with a reduction of velocity of 10 per cent 
= 1225—312=913 ft.-lbs. Let ^'l = maximum velocity of fly- 
wheel rim; 7;2= minimum velocity of fly-wheel rim; TF = weight 
of the fly-wheel rim. The energy it is capable of giving out, 

while its velocity is reduced from Vi to V2, = , and 

the value of W must be such that this energy given out shall 
equal 913 ft.-lbs. Hence the following equation may be written: 

— ^ — = 01^. 



FLY -IV HEELS. 261 



Therefore lF = 2il^f 

The punch-shaft makes 30 revolutions per minute and the pulley- 
shaft 30X5 =150 ^AT revolutions per minute. Hence Vi in 

ND- 
feet per second^— 7 — , D being fly-wheel diameter in feet =3 feet. 

150X37: 

^i=-V^- =23.56; 

'^2= 0.902^1 =21.2; 

^1^ = 555; ^2^=449; ^^i2-^^2^ = io6. 
Hence w ^^-^^^^ ^ SSS ^^s. 

To proportion the rim: A cubic inch of cast iron weighs 

0.26 lb.; hence there must be 7 = 21^^ cu. ins. The cubic 

' 0.26 ^^ 

contents of the rim =mean diameter X tt X its cross-sectional area 

A =2135 cu. ins.; hence 

2135 



If the cross-section were made square its side would =v 18.45 

= 4.3- 

177. Pump Fly-wheel. — The belt for the pump, p. 245, is 

designed for the average work. A fly -wheel is necessar}^ to 
adapt the varying resistance to the capacity of the belt. The 
rate of doing work on the return stroke (supposing no resistance 
due to suction) is only equal to the frictional resistance of the 
machine. During the working stroke the rate of doing work 
varies because the velocity of the plunger varies, although the 
pressure is constant. The rate of doing work is a maximum 
when the velocity of the plunger is greatest. In Fig. 189, A 
is the velocity diagram, B is the force diagram, C is the 



262 



MACHINE DESIGN. 



tangential diagram drawn as indicated on pp. 68-70. The 
belt, 4 inches wide, is capable of applying a tangential force 
of 148 lbs. to the 18-inch pulley- rim. The velocity of the pulley- 
rim =711.5X300=1414'. The velocity of the crank-pin axis 
= 71X0.833 X 50 = 1 3o.9^ Therefore the force of 148 lbs. at 



the pulley- rim corresponds to a force =i48X 



1414 
130.9 



1599 lbs. 



applied tangentially at the crank-pin axis. This may be plotted 
as an ordinate upon the tangential diagram C, from the base 





Fig. 189. 



line XXi, using the same force scale. Through the upper 
extremity of this ordinate draw the horizontal line DE. The 
area between DE and XXi represents the work the belt is 
capable of doing during the working stroke. During the return 
stroke it is capable of doing the same amount of work. But 
this work must now be absorbed in accelerating the fly-wheel. 
Suppose the plunger to be moving in the direction shown by 
the arrow. From E to F the effort is in excess and the fly-wheel 
is storing energy. From F to G the resistance is in excess and 
the fly-wheel is giving out energy. The work the fly-wheel must 
be capable of giving out with the allowable reduction of velocity 
is that represented by the area under the curve above the line 
FG. From G to D, and during the entire return stroke, the 
belt is doing work to accelerate the fly-wheel. This work 



FLY-IVHEELS. 263 

becomes stored kinetic energy in the fly-wheel. Obviously the 
following equation of areas may be written : 

XiEF+ XGD + XHKX 1 = GMF. 

The left-hand member of this equation represents the work 
done by the belt in accelerating the fly-wheel; the right-hand 
member represents the work given out by the fly-wheel to help 
the belt. 

The work in foot-pounds represented by the area GMF 
may be equated with the difference of kinetic energy of the 
fly-wheel at maximum and minimum velocities. To find the 
value of this work: One inch of ordinate on the force diagram 
represents 8520 lbs.; one inch of abscissa represents 0.449 foot. 
Therefore one square inch of area represents 8520 lbs. Xo. 449' 
= 3825.48 ft. -lbs. The area GMF^o.^ sq. ins. Therefore 
the work = 3825.48X0.4 = 1530 ft.-lbs. =£. The difference of 

W 

kinetic QnQ.xgy =— {v)i^ —v^) ^i^^o\ W equals the weight of 

the fly-wheel rim. Hence 

n^_ ^53oX32-2X2 

Assume the mean fly-wheel diameter = 2.5 feet. It will be 
keyed to the pulley-shaft, and will run 300 revolutions per 
minute, = 5 revolutions per second. The maximum velocity 
of fly-wheel rim =7:X2.5 X5 =39.27 =2^1. Assume an allowable 
variation of velocity, =5 per cent. Then x^2 = 37-27 X0.95 = 
37.3; 1/12=1542.3; 1/2 = 1391-3; ^^1^ -1^22=151- Hence 

^^ 1530X32.2X2 ^^ ^^^^ 
151 

There must be 651 ^0.26 cu. in. in the rim, =2504, The mean 
circumference =30" X;r= 94.2". Hence the cross-sectional area 



264 MACHINE DESIGN. 

of rim =2504-^94.2 =26.6 sq. ins. The rim may be made 
4.5" X 6". 

The frictional resistance of the machine is neglected. It 
might have been estimated and introduced into the problem as a 
constant resistance. 

178. Steam-engine Fly-wheel. — From given data draw the 
indicator-card as modified by the acceleration of reciprocating 
parts. See page 67 and Fig. 46. From this and the velocity 
diagram construct the diagram of tangential driving force, 
Fig. 47. Measure the area of this diagram and draw the equiva- 
lent rectangle on the same base. This rectangle represents the 
energy of the uniform resistance during one stroke; while the 
tangential diagram represents the work done by the steam upon 
the crank-pin. The area of the tangential diagram which 
extends above the rectangle represents the work to be absorbed 
by the fiy-wheel with the allowable variation of velocity.* Find 
the value of this in foot-pounds, and equate it to the expression 
for difference of kinetic energy at maximum and minimum 
velocity. Solve for W, the weight of fly-wheel. 

179. Stresses in Fly-wheel Rims. — Mathematical analyses 
of the stresses in fly-wheel rims are unsatisfactory. In the first 
place, in order to get solutions of reasonable simplicity it is 
customary to make assumptions which are contrary to the 
actual conditions; and in the second place, no satisfactory 
data exist concerning the strength of cast iron in such heavy 
sections as are used in large engine fly-wheels. An examination 
of the nature of the stresses, however, will indicate the points 
to be looked out for in design. 

Considering a ring of hollow cylindrical form, comparatively 

* For compound engines and for varying resistances the diagrams should be 
constructed for the complete cycle. For full treatment of the problem of fly 
wheels for engines driving alternators the reader is referred to the Trans. A. S. 
M. E., Vol. XXII, p. 955, and Vol. XXIV, p. 98. 



FLY-IVHEELS. 



26s 




thin radially, it can be showTi that, when it is rotated about its 
axis, tension is set up in the ring proportional to the weight of 
the material used and the square of the linear velocity. This 
tension is due solely to the action of "centrifugal force" and is 
termed "centrifugal tension." 

Consider the half- ring shown in Fig. 190: 

z;=the velocity of the rim in feet per second, 
c = " centrifugal force" per foot of rim; 
i^= radius in feet; 
A =area of rim in square inches; 
P= total tension in rim in pounds; 
jt =unit tensile stress in rim in pounds; 
w = weight of material as represented by a piece i inch 

square and i foot long; 
g =32.2 feet per second per second; 
2P=sum of horizontal components of all the small centri- 
fugal forces cds\ 
Each horizontal component =cds cos (9, w^hich may be written 
cR cos d dd, because ds=Rdd. 



Fig. 190. 



2P 



/; 



cR cos 



>cR\ 



:. P-=cR. 



But 



R ~ g R 



P = 



Wv^ 



W beino: the w^eight of one linear foot of rim =^wA, 



i66 MACHINE DESIGN. 

Also, P=M; 



r.UA- 


g 


wAv^ 


■•■ Ir- 


g ' 





For cast iron, putting /< =20,000 lbs. the ultimate strength, 
and w =0.26X12 lbs., it follows that 1^=454 feet per second. 
In other words a cast-iron ring will burst at a speed of 454 feet 
per second. Furthermore, an examination of the formula 
shows that for a ring this bursting velocity depends not at all 
on the size or shape of the cross-section but only on the material 
used as represented by ft and w. 

This centrifugal tension causes a corresponding elongation 
of the material and therefore an increase in the radius of the 
ring. A free ring of whatever cross-section can and does take 
the new radius and the tension on all sections =ft pounds per 
square inch. 

With the introduction of rigidly fastened arms a number of 
new and vital elements enter into the problem. An arm of the 
same original length as the original radius of the rim when 
rotated about an axis perpendicular to its inner end will also 
suffer an elongation due to centrifugal action. The amount of this 
radial elongation will vary with the form of the arm, but in no 
practical case will it amount to as much as one third of the 
radial increase of the ring rotating at the same speed. 

To accommodate this difference the arm, if rigidly fastened 
to hub and rim, will be extended lengthwise by the rim and the 
rim will be drawn in, out of its regular circular form, by the 
arm. The relation between the amount the arm is dra^vn out 
and the amount the rim is drawn in is governed by the propor- 
tions of these parts. 



FLY-IVHEELS. 267 

The result is that the rim tends to bow out between the arms 
and really become akin to a uniformly loaded continuous beam 
with the dangerous sections midway between the arms and at 
the points of junction of arms and rim. The fallacy of applying 
the ring theory solely to the fly-wheel rim becomes evident at 
once. In a free ring the form of cross-section is immaterial, as 
the section is subjected only to tension. In the rim with arms 
the form of cross-section becomes a vital point, as the rim is 
subjected to flexure as well as tension, and the strength of a 
member to resist flexure depends directly upon the modulus 
of the section. 

In addition to the foregoing stresses, which are induced 
under all conditions, even under the extreme supposition that 
the wheel is rotating at a perfectly uniform rate, there are others 
when the rim is considered as performing its functions — i.e., 
in a balance-wheel, absorbing or giving out energy by changes 
of velocity and, in a band-wheel, transmitting the power. 

This may be seen by reference to Fig. 191. A shows the 
relation between rim, arm, and hub when the wheel is at rest 
or rotating uniformly and not transmitting 
any power. B shows the relation when 
work is being done. The arm becomes an 
encastre beam and corresponding stresses 
are induced in it. Furthermore, the bend- 
ing of the arm tends to shorten it radially, thus drawing in the 
outer end, which increases the flexure in the rim. In addition 
to the foregoing there are stresses in the rim due to the weight 
of the wheel, shrinkage, etc., which cannot be eliminated. 

180. Stresses in Arms of Pulleys or Fly-wheels. — The arms 
are principally stressed by the bending moment due to variations 
of velocity of the wheel or to the power transmitted. 

Let r = the greatest turning moment transmitted in inch- 
pounds; 




26S MACHINE DESIGN. 

w= number of arms; 

}t=ssiie unit stress in outer fiber of arm in pounds per 
square inch; 

— = modulus of section of arm, dimensions in inches. 
c 

Then 

T=nft— may be written and solved for—. 
c c 

Having determined upon the form of cross-section the dimensions 

can be determined from this value of—. 

c 

If T is unknown the arms can be made as strong as the shaft 

l)y equating the twisting strength of the shaft to the bending 

strength of the arms, thus: 

Trr^ I 

Jfs= allowable shearing stress in outer fiber of shaft, pounds per 

square inch; 
r = radius of shaft in inches; 

#n, jt, and — as before. 
Consider junction of arm and hub next. (See 
Fig. 192.) 
Fig. 192. 

The tendency for the arm to fail through 

flexure on the section A-A may be equated to the tendency 
for the bolts 2 and 3 to shear off, using i as a pivot. 

Let A = combined shearing areas of 2 and 3, square inches; 
/8= allowable shearing stress of 2 and 3 in pounds per 

square inch; 
/=distance between centers i and 2, and i and 3, in 
inches; 



Then 



FLY-WHEELS. 269 

/<= allowable stress in outer fiber of arm in pounds per 
square inch; 

— = modulus of arm section, dimensions in inches. 
c 



AUi=h^. 



which can be solved for A, the desired area. 

If the arm is bolted to the run a similar method may be 
employed to make the bolts as strong as the arm. 

181. Construction of Fly-v/heels. — Since weight is so great 
a factor in fly-wheels it has been common practice to make 
them of that material which combines greatest weight with 
least cost, namely, cast iron. That this is not always safe 
practice has been conclusively demonstrated by many serious 
accidents. 

Up to 10 feet in diameter the wheels are generally cast 
in a single piece. Occasionally the hub is divided to relieve 
the stresses due to cooling. In such cases, supposing the wheel 
to have six arms, the hub is made in three sections, each having 
a pair of arms nmning to the rim. Since the sections are inde- 
pendent, any pair of arms can adjust itself to the conditions cf 
shrinkage without subjecting the other arms to indeterminate 
stresses. The hub sections are separated from each other by 
a space of half an inch or less and this is filled with lead or 
babbitt metal. Then shrink-rings or bolts are used to hold 
the sections together. Sometimes the hub is only split into two 
parts. 

For reasons connected chiefly with transportation, wheels 
from 10 to 15 feet in diameter are cast in two halves which 
are afterwards joined together by flanges and bolts at the rim, 
and shrink-rings or bolts at the hub. 



2 70 MACHINE DESIGN. 

In still larger and heavier wheels the hub is generally made 
entirely separate from the arms. The rim is made in as many 
segments as there are arms. Sometimes the arm is cast with 
the segment and sometimes the arms and segments are cast 
separately. The hub is commonly made in the form of a pair 
of disks having a space between them to receive the arms which 
are fastened to them by means of accurately fitted through bolts. 

Unless the wheel is to be a forced fit on its shaft it is best 
to have three equally spaced keyways, so that it may be kept 
accurately centered with the shaft. 

In these large wheels the joints of the segments of the rim 
are usually midway between the arms and steel straps or links 

CT^~'~S~~^ "T^C^^)^ ^^^^ ^^ ^^^ shown in Fig. 193 are 
\ — i — --/ W-""^""^^ heated and dropped into recesses pre- 
•^9^' viously fitted to receive them. As they 

cool, their contraction draws the joint together. They should 
not, however, be subjected to a very great initial tension of this 
sort. The form shown at A is most commonly used. The links 
are made of high-grade steel and their area is such that their ten- 
sile strength equals that of the reduced section of the rim. The 
areas subjected to shear and compression must also have this 
strength. 

Taking the nature of the stresses into consideration it is clear 
that the rim should always be as deep radially as possible to 
resist the flexure action, also that the arms should be near together. 
Many arms are much better than a few and a disk or web is still 
better.* 

The strongest wheel having arms will be one whose rim is 
cast in a single piece, while the arms and hubs are cast as a second 
piece. On the inside of the rim there are lugs between which 



* Disk wheels have the further advantage of offering less resistance to the 
air. This maybe a considerable item. See Cassier's Mag., Vol. 23, pp. 577 and 
761. 




FLY -IVH EELS. 271 

the ends of the arms fit so that there is a space of about one 
fourth inch all around. (See Fig. 194.) This space may be 
filled with oakum well driven in. It is clear that the rim in this 
case acts as a free ring and is subjected solely to centrifugal 
tension.* 

Joints in the rim must always be a source of weakness whether 
located at the end of the arms or midway between arms. 

^^^^ 

Tension 
Rod 

Fig. 194. Fig. 195. 

If a flanged joint midway between the arms is used, such as 
is showTi in Fig. 195, which is common practice for split band- 
wheels of medium size, the flanges should be deep radially and 
well braced by ribs. The bolts should be as close to the rim 
as possible, and a tension rod should carry the extra stress (due 
to the w^eight of the hcaw joint and its velocity) to the hub. 
Experiments made by Prof. C. H. Benjamin f show that the use 
of such tie-rods increases the strength of the wheel 100 per cent 
over that of a similar wheel without tie-rods. He also found 
that jointed rims are only one fourth as strong as solid rims. 

Probably as strong a form of cast-iron built-up wheel for 
heavy duty as any yet designed is one described by Mr. John 
Fritz,J having a hollow rim and many arms. 

But, at the best, cast iron is an uncertain material to use 
for such tensile and flexure stresses as are induced in a heavy- 
duty fly-wheel, and it is wiser to make such wheels of structural 
steel. A built-up wheel having a disk or web of steel plates and 
a rim of the same material, all joints being carefully "broken '^ 
and strongly riveted, is so much better than any built-up cast- 

* See Trans. A. S. M. E., Vol. XX, p. 944, and Voi. XXI, p. 322. 
t Ibid., Vols. XX and XXIII. 
X Ibid., Vol. XXI. 



2 7 2 MACHINE-DESIGN. 

iron wheel that the latter are passing out of use. The steel 
wheels can have at least twice the rim velocity of the cast wheels 
with greater safety and may therefore be much lighter for the 
same duty. Their lesser weight makes less pressure on the 
bearings and consequently less friction loss.* 

Plate II shows forms of rim joints for split rim flywheels 
and pulleys which are probably as strong as any that can be 
devised. They are taken from the American Machinist, Vol. 30. 
It is a mistake, however, to believe that any of these joints will 
will give as strong a wheel as one having a solid rim. 

* For drawings and descriptions of wheels made of forged materials the reader 
is referred to Vol. XVII, Trans. A. S. M. E., and Power, April 1894, Nov. 1895^ 
Jan. 1896, and Nov. 1897. 



Plate II. 




HAIGHT'S JOINT FOR HEAVY RIM. 



FLANGED JOINT OVER-ARM 
SEGMENTAL RIM, 



c3(^^(6) o 




DOUBLE ARM JOINT FOR WIDE RIM. 




E[TF 




\o u o u o^ 



DOUBLE ARM JOINT FOR SHEAVE WHEEL. 



^^n 



CHAPTER XVII. 

TOOTHED WHEELS OR GEARS. 

182. Fundamental Theory of Gear Transmission.^ — When 
toothed wheels are used to communicate motion, the motion 
elements are the tooth surfaces. The contact of these surfaces 
with each other is Une contact. Such pairs of motion elements 
are called higher pairs, to distinguish them from lower pairs, 
which are in contact throughout their entire surface. Fig. 196 
shows the simplest toothed-wheel mechanism. There are three 
Hnks, a, b, and c, and therefore three centros, ab, be, and ac. These 
centros must, as heretofore explained, lie in the same straight 
Hne. ac and ab are the centers of the turning pairs connecting 
c and b to a. It is required to locate be on the line of centers. 

When the gear e is caused to rotate uniformly with a certain 

angular velocity, i.e., at the rate of m revolutions per minute, 

it is required to cause the gear b to rotate uniformly at a rate 

of n revolutions per minute. The angular velocity ratio is there- 

tn 
fore constant and = — . The centro be is a point on the line of 

centers which has the same linear velocity whether it is con- 
sidered as a poiQt in 6 or c. The Knear velocity of this point be 
mb = 2nRin; and the linear velocity of the same point 'me = 27zR2m ; 
in which Ri = radius of be in 6, and i?2= radius of be in e. But 
this linear velocity must be the same in both cases, and hence the 
above expressions may be equated thus: 

2nR\n = 27zR2fn, 

275 



276 

whence 



MACHINE DESIGN. 

Ri m 
R2^n' 



Hence be is located by dividing the Hne of centers into parts which 

are to each other inversely as the angular velocities of the gears. 

Thus, let ab and ac, Fig. 197, be the centers of a pair of gears 

whose angular velocity ratio =—. Draw the line of centers; 

divide it into m-hn equal parts; m of these from ab toward the 
right, or n from ac toward the left, will locate be. Draw circles 
through be, with ab and ae as centers. These circles are the 
centrodes of be and are called piteh cireles. It has been aheady 
explained that any motion may be reproduced by rolHng the 
centrodes of that motion upon each other without slipping. 




Fig. 196. 



Fig. 197. 



Therefore the motion of gears is the same as that which w^ould 
result from the rolling together of the pitch circles (or cylinders) 
without shpping. In fact, these pitch cylinders themselves 
might be, and sometimes are, used for transmitting motion of 
rotation. Shpping, however, is apt to occur, and hence these 
"friction-gears" cannot be used if no variation from the given 
velocity ratio is allowable. Hence teeth are formed on the 
wheels which engage with each other, to prevent sHpping. 

183. Definitions. — If the pitch circle be divided into as many 
equal parts as there are teeth in the gear, the arc included between 



TOOTHED IVHEELS OR GEARS. 277 

two of these divisions is the circular pitch * of the gear. Circular 
pitch may also be defined as the distance on the pitch circle 
occupied by a tooth and a space; or, otherwise, it is the distance 
on the pitch circle from any point of a tooth to the corresponding 
point in the next tooth. A fractional tooth is impossible, and 
therefore the circular pitch must be such a value that the pitch 
circumference is divisible by it. Let P= circular pitch in inches; 
let D= pitch diameter in inches; A^=number of teeth; then 

NP=7:D; N=^; D ; P=^. From these relations 

any one of the three values, P, D, and N, may be found if the 

other two are given. 

Diametral pitch is the number of teeth per inch of pitch 

N 
diameter. Thus if ^ = diametral pitch, p=j^- Multiplying the 

^ 7zD ^ N ^ . ^ nD N 

two expressions, P=~^t and p =-7t, together gives Pp =-rjr .jr=Tc. 

Or, the product of diametral and circular pitch =7r. Circular 
pitch is usually used for large cast gears, and for mortise-gears 
(gears with wooden teeth inserted). Diametral pitch is usually 
used for small cut gears. 

In Fig. 198, h, c, and k are pitch points of the teeth; the arc 
hk is the circular pitch ; ah is the jace of the tooth ; hm is the -flank 
of the tooth ; the whole curve ahm is the pro-file of 
the tooth; AD h the total depth of the tooth; AC 
is the working depth] AB h the addendum; a 
circle through^ is the addendum circle. Clear- 
ance is the excess of total depth over working ^^^' 
depth, ^CD. Backlash is the width of space on the pitch line 
minus the width of the tooth on the same line. In cast gears 
whose tooth surfaces are not "tooled," backlash needs to be 




Sometimes called circumferential -pitch. 



278 MACHINE DESIGN. 

allowed, because of unavoidable imperfections in the surfaces. 
In cut gears, however, it rnay be reduced almost to zero, and the 
tooth and space, measured on the pitch circle, may be considered 
equal. 

184. Conditions Governing Forms of Teeth. — Teeth of almost 
any form may be used, and the average velocity will be right. 
But if the forms are not correct there will be continual variations 
of velocity ratio between a minimum and maximum value. These 
variations are in many cases unallowable, and in all cases unde- 
sirable. It is necessary therefore to study tooth outlines which 
shall serve for the transmission of a constant velocity ratio. 

The centro of relative motion of the two gears must remain 
in a constant position in order that the velocity ratio shall be 
constant. The essential condition jor constant velocity ratio is, 
therefore, that the position of the centro oj relative motion of the 
gears shall remain unchanged. If A and B, Fig. 199, are tooth 
surfaces in contact at a, their only possible relative motion, if 
they remain in contact, is slipping motion along the tangent CD. 
The centro of this motion must be in EF, a normal to the tooth 
surfaces at the point of contact. If these be supposed to be 
teeth of a pair of gears, b and c, whose required velocity ratio 
is known, and whose centro, be, is therefore located, then in 
order that the motion communicated from one gear to the other 
through the point of contact, a, shall be the required motion, it 
is necessary that the centro of the relative motion of the teeth 
shall coincide with be. 

185. Illustration. — In Fig. 200, let ac and ab be centers of 
rotation of bodies b and c, and the required velocity ratio is such 
that the centro of b and c falls at be. Contact between b and c 
is at p. The only possible relative motion if these surfaces re- 
main in contact is slipping along CD; hence the centro of this 
motion must be on EF, the normal to the tooth surfaces at the 
point of contact. But it must also be on the same straight Une 



TOOTHED WHEELS OR GEARS. 279 

with ac and ah\ hence it is at he, and the motion transmitted for 
the instant, at the point p, is the required motion, because its 
centro is at he. But the curves touching at p might be of such 
form that their common normal at p would intersect the Une of 
centers at some other point, as K, which would then become the 
centro of the motion of h and c for the instant, and would corre- 
spond to the transmission of a different motion. The essential 
condition to be fulfilled by tooth outlines, in order that a con- 
stant velocity ratio may be maintained, may therefore be stated 
as follows: The tooth outlines must he such that their normal at 
the point oj contact shall always pass through the centro corre- 
sponding to the required velocity ratio. 

Fig. 199. 




Fig. 200. 

186. Given Tooth Outline to Find Form of Engaging Tooth. 

— Having given any curve that will serve for a tooth outline 

in one gear, the corresponding curve may be found in the other 

gear, which will engage with the given curve and transmit a 

m 
constant velocity ratio. Let — be the ejiven velocity ratio. 

n ^ 

w + w = the sum of the radii of the two gears. Draw the line 

of centers AB, Fig. 201. Let P be the "pitch point," i.e., the 

point of contact of the pitch circles or the centro of relative 

motion of the two gears. To the right from P lay off a distance 

PB=m; from P toward the left lay off PA =n. A and B will 

then be the required centers of the wheels, and the pitch circles 



28o 



MACHINE DESIGN. 



may be drawn through P. Let ahc be any given curve on the 
wheel A. It is required to find the curve in B which shall engage 
with ahc to transmit the constant velocity ratio required. A 
normal to the point of contact must pass through the centro. 
Tf, therefore, any point, as a, be taken in the given curve, and a 
normal to the curve at that point be drawn, as aa, then when a 
is the point of contact, a will coincide with P. Also, if cy is a 
normal to the curve at c, then ;- will coincide with P when c is 
the point of contact between the gears; and since 6 is in the pitch 
hne, it will itself coincide with P when it is the point of contact. 

Fig. 201. 




Fig. 202. 

Since the two pitch circles must roll upon each other without 
slipping, it follows that the arc Pct'=arcPQf, arc P6'=arc Ph^ 
and arc Pf = arc Py. 

Rotate the point a, about A, through the angle 6. At the 
same time a' rotates backward about B through the angle 6^ 
and a and a' coincide at P. Pa" represents the rotated position 
of the normal aa. Rotate Pa'^ about B through the angle (9'; 
P will coincide with a' and a'^ will locate the point a' of the desired 
tooth outline of gear B. The point h' of the desired outline is 
readily located by merely laying off arc Ph' =arc Ph. 

c' is located by the same method we employed to determine 
a' . This will give three points in the required curve, and through 



TOOTHED fVHEELS OR GEARS. 281 

these the curve may be drawn. The curve could, of course, be 
more accurately determined by using more points. 

Many curves could be drawn that would not serve for tooth 
outlines; but, given any curve that will serve, the corresponding 
curve may be found. There would be, therefore, almost an 
infinite number of curves that would fulfill the requirements of 
correct tooth outlines. But in practice two kinds of curves are 
found so convenient that they are most commonly, though not 
exclusively, used. They are cycloidal and involute curves. 

187. Cycloidal Tooth Outlines. — It is assumed that the char- 
acter of cycloidal curves and method of drawing them is under- 
stood. 

In Fig. 202, let h and c be the pitch circles of a pair of wheels, 
always in contact at he. Also, let m be the describing circle in 
contact with both at the same point. M is the describing point. 
When one curve rolls upon another, the centro of their relative 
motion is always their point of contact. For, since the motion 
of rolling excludes slipping, the two bodies must be stationary, 
relative to each other, at their point of contact; and bodies that 
move relative to each other can have but one such stationary 
point in common — their centro. When, therefore, m rolls in 
or upon h or c, its centro relatively to either is their point of con- 
tact. The point Af, therefore, must describe curves whose 
direction at any point is at right angles to a line joining that 
point to the point of contact of m with the pitch circles. Suppose 
the two circles h and c to revolve about their centers, being always 
in contact at hc] suppose m to rotate at the same time about its 
center, the three circles being always in contact at one point and 
having no slip. The point M will then describe simultaneously 
a curve, 6', on the plane of h, and a curve, c', on the plane of c. 
Since M describes the curves simultaneously, it w^ill always be 
the point of contact between them in a,ny position. And since 
the point M moves always at right angles to a Hne which joins it 



282 MACHINE DESIGN. 

to hCj therefore the normal to the tooth surfaces at their point 
of contact will always pass through he, and the condition for 
constant velocity ratio transmission is fulfilled. But these curves 
are precisely the epicycloid and hypocycloid that would be drawn 
by the point M in the generating circle, by rolling on the out- 
side of h and inside of c. Obviously, then, the epicycloids and 
hypocycloids generated in this way, used as tooth profiles, will 
transmit a constant velocity ratio. 

This proof is independent of the size of the generating circle, 
and its diameter may therefore equal the radius of c. Then the 
hypocycloids generated by rolhng within c would be straight 
lines coinciding with the radius of c. In this case the flanks 
of the teeth of c become radial lines, and therefore the teeth are 
thinner at the base than at the pitch line; for this reason they 
are weaker than if a smaller generating circle had been used. All 
tooth curves generated with the same generating circle will work 
together, the pitch being the same. It is therefore necessary 
to use the same generating circle for a set of gears which need 
to interchange.^ 

The describing circle may be made still larger. In the first 
case the curves described have their convexity in the same direction, 
i.e., they lie on the same side of a common tangent. When the 
diameter of the describing circle is made equal to the radius 
of c, one curve becomes a straight-hne tangent to the other curve. 
As the describing circle becomes still larger, the curves have their 
convexity in opposite directions. As the circle approximates 
equality with c, the hypocycloid in c grows shorter, and finally 
when the describing circle equals c, it becomes a point which is 
the generating point in c, which is now the generating circle. If 
this point could be replaced by a pin having no sensible diameter, 
it would engage with the epicycloid generated by it in the other 
gear to transmit a constant velocity ratio. But a pin without 

* See § 192. 




TOOTHED IVHEELS OR GEARS. 283 

sensible diameter will not serve as a wheel-tooth, and a proper 
diameter must be assumed, and a new curve laid off to engage 
with it in the other gear. In Fig. 203, AB is the epicycloid gener- 
ated by a point in the circumference of the 
other pitch circle. CD is the new curve 
drawn tangent to a series of positions of the 
pin as shown. The pin will engage with 
this curve, CD, and transmit the constant 
velocity ratio as required. In Fig. 202, let it ^ig 

be supposed that when the three circles rotate 
constantly tangent to each other at the pitch point he, a pencil is 
fastened at the point M in the circumference of the describing 
circle. If this pencil be supposed to mark simultaneously upon 
the planes of 6, c, and that of the paper, it will describe upon h 
an epicycloid, on c a hypocycloid, and on the plane of the 
paper an arc of the describing circle. Since M is always 
the point of contact of the cycloidal curves (because it gen- 
erates them simultaneously), therefore, in cycloidal gear-teeth, 
the locus or path of the point of contact is an arc of the describ- 
ing circle. The ends of this path in any given case are located 
by the points at which the addendum circles cut the describing 
circles. 

In the cases already considered, where an epicycloid in one 
wheel engages with a hypocycloid in the other, the contact of 
the teeth with each other is all on one side of the hne of centers. 
Thus, in Fig. 202, if the motion be reversed, the curves will be 
in contact until M returns to be along the arc MD-bc; but after 
M passes be contact wall cease. If c were the driving-wheel, the 
point of contact would approach the Hne of centers ; if b were the 
driving-wheel the point of contact would recede from the line of 
centers. Experience shows that the latter gives smoother nmning 
because of better conditions as regards friction between the tooth 
surfaces. It would be desirable, therefore, that the wheel with 
the epicycloidal curves should always be the driver. But it 



284 MACHINE DESIGN. 

should be possible to use either wheel as driver to meet the varying 
conditions in practice. 

Another reason why contact should not be all on one side of 
the line of centers may be explained as follows. 

i88. Definitions : Pitch-arc, Arc of Action, Line of Pressure. 
— The angle through which a gear-wheel turns w^hile one of its 
teeth is in contact with the corresponding tooth in the other gear 
is called the angle of action. It is found by drawing radial lines 
from the center of the pitch circle to the two ends of the path of 
action. The arc of the pitch circle corresponding to the angle 
of action is called the arc of action. 

The arc of action must be greater than the "pitch arc^' (the 
arc of the pitch circle that includes one tooth and one space), or 
else contact will cease between one pair of teeth before it begins 
between the next pair. Constrainment would therefore not be 
complete, and irregular velocity ratio with noisy action would 
result. 

In Fig. 204, let AB and CD be the pitch circles of a pair of 
gears and E the describing circle. Let an arc of action be laid 
off on each of the circles from P, as Pa, Pc, and Pe. Through 
e, about the center O, draw an addendum circle; i.e., the circle 
which Hmits the points of the teeth. Since the circle E is the 
path of the point of contact, and since the addendum circle 
Hmits the points of the teeth, their intersection, e, is the point at 
which contact ceases, rotation being as indicated by the arrow. 
If the pitch arc just equals the assumed arc of action, contact 
will be just beginning at P when it ceases at e; but if the pitch 
arc be greater than the arc of action, contact will not begin at 
P till after it has ceased at e, and there will be an interval when 
AB will not drive CD. The greater the arc of action the greater 
the distance of e from P on the circumference of the describing 
circle. The direction of pressure between the teeth is always 
a normal to the tooth surface, and this always passes through 



TOOTHED IVHEELS OR GEARS. 



285 



the pitch point; therefore the greater the arc of action — i.e.-, 
the greater the distance of e from P — the greater the obUquity 
of the line of pressure. The pressure may be resolved into two 
components, one at right angles to the line of centers and the 
other parallel to it. The first is resisted by the teeth of the follower- 
wheel, and therefore produces pressure with resulting friction. 
Hence it follows that the greater the arc of action the greater 
will be the average obUquity of the line of pressure, and there- 
fore the greater the component of the pressure that produces 





Fig. 204. 



Fig. 205. 



wasteful friction. If it can be arranged so that the arc of action 
shall be partly on each side of the line of centers, the arc of action 
may be made greater than the pitch arc (usually equal to about 
I J times the pitch arc); then the obliquity of the pressure-line 
may be kept w^ithin reasonable limits, contact betw^een the teeth 
will be insured in all positions, and either wheel may be the 
driver. This is accomplished by using two describing circles as 
in Fig. 205. Suppose the four circles A, B, a, and ,5 to roll con- 
stantly tangent at P. a w^ill describe an epicycloid on the plane 
of B, and a hypocycloid on the plane of A. These curves wall 
engage with each other to drive correctly. ^ \yi\\ describe an 
epicycloid on A, and a hypocycloid on B. These curves will 
engage, also, to drive correctly. If the epicycloid and hypocycloid 
in each gear be drawn through the same point on the pitch circle, 
a double curve tooth outline will be located, and one curve will 



286 MACHINE DESIGN. 

engage on one side of the line of centers and the other on the 
other side. If A drives as indicated by the arrow, contact will 
begin at Z>, and the point of contact will follow an arc of a to P, 
and then an arc of /? to C 

189. Involute Tooth Outlines. — If a string is wound around a 
cylinder and a pencil-point attached to its end, this point will 
trace an involute on a plane normal to the axis of the cylinder 
as the string is unwound from the cylinder. Or, if the point 
be constrained to follow a tangeiit to the cylinder, and the string 
be unwound by rotating the cylinder about its axis, the point 
will trace an involute on a plane that rotates with the cylinder. 

Illustration. — ^Let a, Fig. 206, be a circular piece of wood free 
to rotate about C; /? is a circular piece of cardboard made fast 
to a; ^J5 is a straight-edge held on the circumference of a, 
having a pencil-point at B, As B moves along the straight-edge 
io A, a and /? rotate about C, and B traces an involute DB upon 
|9, the relative motion of the tracing point and ^ being just the 
same as if the string had been simply unwound from a fixed. 
If the tracing point is caused to return along the straight-edge 
it will trace the involute BD in a reverse direction. 





Fig. 206. Fig. 207. 

The centro of the tracing point is always the point of tan- 
gen cy of the string with the cylinder; therefore the string, or 
straight-edge, in Fig. 208, is always at right angles to the direc- 
tion of motion of the tracing point, and hence is always a normal 
to the involute curve. Let a and 3, Fig. 207, be two base cylin- 
ders; let AB be a cord wound upon a and /? and passing through 



TOOTHED IVHEELS OR GEARS. 287 

the centre P, which corresponds to the required velocity ratio. 
Let a and ^ be supposed to rotate so that the cord is wound from 
/? upon a. Then any point in the cord will move from A toward 
B, and, if it be a tracing-point, will trace an involute of a on the 
plane of a (extended beyond the base cylinder) , and will also trace 
an involute of /? upon the plane of /?. These two involutes will 
serve for tooth profiles for the transmission of the required con- 
stant velocity ratio, because AB is the constant normal to both 
curves at their point of contact, and it passes through P, the 
centro corresponding to the required velocity ratio. Hence the 
necessary condition is fulfilled. The pitch circles will have OP 
and O'P as their respective radii. 

Since a point in the hne AB describes the two involute curves 
simultaneously, the point of contact of the curves is always in 
the line AB. And hence AB \?> the path of the point of contact. 
In any given case the two ends of the path lie at the intersections 
of the addendum circles with AB. The angle oj action and the 
arc oj action are found by drawing radial lines from the center 
of the pitch circle to the ends of the path of contact. 

One of the advantages of involute curves for tooth profiles 
is that a change in distance between centers of the gears does 
not interfere with the transmission of a constant velocity ratio. 
This may be proved as follows: In Fig. 207, from similar triangles 

-:rr-r- = TTTF, '•, that is, the ratio of the radii of the base circles (i.e., 
O'A OP ^ ' 

sections of the base cylinders) is equal to the ratio of the radii of 

the pitch circles. This ratio equals the inverse ratio of angular 

velocities of the gears. Suppose now that O and O' be moved 

nearer together; the pitch circles will be smaller, but the ratio 

of their radii must be equal to the unchanged ratio of the 

radii of the base circles, and therefore the velocity ratio remains 

unchanged. Also the involute curves, since they are generated 

from the same base cylinders, will be the same as before, and 



288 MACHINE DESIGN. 

therefore, with the same tooth outHnes, the same constant velocity 
ratio will be transmitted as before. 

190. Racks. — A rack is a wheel whose pitch radius is infinite ; 
its pitch circle, therefore, becomes a straight line, and is tangent 
to the pitch circle of the wheel, or pinion,* with which the rack 
engages. The line of centers is a normal to the pitch line of the 
rack, through the center of the pitch circle of the pinion. The 
pitch of the rack is determined by laying off the circular pitch 
of the engaging wheel on the pitch line of the rack. The curves 
of the cycloidal rack-teeth, Hke those of wheels of finite radius, 
may be generated by a point in the circumference of a circle which 
rolls on the pitch circle. Since, however, the pitch circle is now 
a straight line, the tooth curves will be cycloids, both for flanks 
and faces. In Fig. 208, ^^ is the pitch circle of the pinion and 
CD is the pitch fine of the rack; a and b are describing circles. 
Suppose, as before, that all move without sKpping and are con- 
stantly tangent at P. A point in the circumference of a will then 





Fig. 208. Fig. 209. 

describe simultaneously a cycloid on CD, and a hypocycloid 
within AB. These will be correct tooth outlines. Also, a point 
in the circumference of b will describe a cycloid on CD and an 
epicycloid on AB. These will be correct tooth outlines. To 
find the path of the point of contact, draw the addendum circle 
EF of the pinion, and the addendum Hne GH of the rack. When 
the pinion turns clockwise and drives the rack, contact will begin 

* Pinion is a word to denote a gear having a low number of teeth, or the smaller 
one of a pair of engaging gears. 



TOOTHED WHEELS OR GE/1RS. 289 

at e and follow arcs of the describing circles through P to 
K. It is obvious that a rack cannot be used where rotation 
is continuous in one direction, but only where motion is 
reversed. 

Involute curves may also be used for the outhnes of rack 
teeth. Let CD and OD\ Fig. 209^ be the pitch lines. When it 
is required to generate involute curves for tooth outhnes, for a 
pair of gears of finite radius, a line is drawn through the pitch 
point at a given angle to the line of centers (usually 75°) ; this 
line is the path of the point which generates two involutes simul- 
taneously, and therefore the path of the point of contact between 
the tooth curves. It is also the common tangent to the two base 
circles, which may now be drawn and used for the describing of 
the involutes. To apply this to the case of a rack and pinion, 
draw EF, Fig. 209, making the desired angle with the line of 
centers, OP. The base circles must be dravvTi tangent to this 
Une; AB will therefore be the base circle for th;^ pinion. But 
the base circle in the rack has an infinite radius, and a circle of 
infinite radius drawn tangent to EF would be a straight line 
coincident with EF. Therefore EF is the base line of the rack. 
But an involute to a base circle of infinite radius is a straight 
line normal to the circumference — in this case a straight line per- 
pendicular to EF. Therefore the tooth profiles of a rack in the 
involute system will always be straight lines perpendicular to the 
path of the describing point, and passing through the pitch points. 
If, in Fig. 209, the pinion move clockwise and drive the rack, the 
contact will begin at £, the intersection of the addendum line 
of the rack GH, and the path of the point of contact EF, and 
will follow the line EF through P to the point where EF cuts the 
addendum circle LM of the pinion. 

191. Annular Gears. — Both centers of a pair of gears may be 
on the same side of the pitch point. This arrangement corre- 
sponds to what is known as an annular gear and pinion. Thus, 



290 



MACHINE DESIGN. 



in Fig. 2io,AB and CD are the pitch circles, and their centers are 
both above the pitch point P. Teeth may be constructed to 
transmit rotation between AB and CD. AB will be an ordinary 





Fig. 2IO. 



Fig. 211. 



spur pinion, but it is obvious that CD becomes a ring of metal 
with teeth on the inside, i.e., it is an annular gear. In this case 
a and /? may be describing circles for cycloidal teeth, and a point 
in the circumference of a will describe hypocycloids simultaneously 
on the planes of AB and CD; and a point in the circumference 
of ^ will describe epicycloids simultaneously on the planes of AB 
and CD. These will engage to transmit a constant velocity 
ratio. Obviously the space inside of an annular gear corresponds 
to a spur- gear of the same pitch and pitch diameter, with tooth 
curves drawn with the same describing circle. Let EF and GHy 
Fig. 2IO, be the addendum circles. If the pinion move clockwise 
driving the annular gear, the path of the point of contact will be 
from e along the circumference of a to P, and from P along the 
circumference of /? to K. 

The construction of involute teeth for an annular gear and 
pinion involves exactly the same principles as in the case of a 
pair of spur- gears. The only difference of detail is that the 
describing point is in the tangent to the base circles produced 
instead of being between the points of tangency. Let O and 0\ 



TOOTHED JVHEELS OR GEy4RS. 291 

Fig. 211, be the centers, and AB and IJ the pitch circles of an 
annular gear and pinion. Through P, the point of tangency of 
the pitch circles, draw the path of the point of contact at the given 
angle with the line of centers. With O and O' as centers draw 
tangent circles to this line. These will be the involute base 
circles. Let the tangent be replaced by a cord, made fast, say, 
at K', winding on the circumference of the base circle CK\ to D^ 
and then around the base circle FE in the direction of the arrow, 
and passing over the pulley G, which holds it in line with PB. 
If rotation be supposed to occur with the two pitch circles always 
tangent at P without sHpping, any point in the cord beyond P 
toward G will describe an involute on the plane //, and another 
on the plane of AB. These will be the correct involute tooth 
profiles required. Draw NQ and LM, the addendum circles. 
Then if the pinion move clockwise, driving the annular gear, 
the point of contact starts from e and moves along the line GH 
through P to K. 

When a pair of spur-gears mesh with each other, the direction 
of rotation is reversed. But an annular gear and pinion meshing 
together rotate in the same direction. 

192. Interchangeable Sets of Gears. — In practice it is desirable 
to have "interchangeable sets " of gears; i.e., sets in which any 
gear will "mesh " correctly with any other, from the smallest 
pinion to the rack, and in which, except for limiting conditions 
of size, any spur-gear will mesh with any annular gear. Inter- 
changeable sets may be made in either the cycloidal or involute 
system. A necessary condition in any set is that the pitch shall 
be constant, because the thickness of tooth on the pitch line must 
always equal the width of the space (less backlash). If this 
condition is unfulfilled they cannot engage, whatever the form 
of the tooth outlines. 

The second condition for an interchangeable set in the 
cycloidal system is that the size oj the describing circle shall be 



292 MACHINE DESIGN. 

constant. If the diameter of the describing circle equal the radius 
of the smallest pinion's pitch circle, the flanks of this pinion's 
teeth will be radial lines, and the tooth will therefore be thinner 
at the base than at the pitch line. As the gears increase in size 
with this constant size of describing circle, the teeth grow thicker 
at the base; hence the weakest teeth are those of the smallest 
pinion. 

It is found unadvisable to make a pinion with less than twelve 
teeth. If the radius of a fifteen-tooth pinion be selected for the 
diameter of the describing circle, the flanks which bound a space 
in a twelve-tooth pinion will be very nearly parallel, and may 
therefore be cut with a milling-cutter. This would not be possi- 
ble if the describing circle were made larger, causing the space 
to become wider at the bottom than at the pitch circle. There- 
fore the maximum describing circle for milled gears is one whose 
diameter equals the pitch radius of a fifteen-tooth pinion and 
it is the one usually selected. Each change in the number of 
teeth with constant pitch causes a change in the size of the pitch 
circle. Hence the form of the tooth outhne, generated by a 
describing circle of constant diameter, also changes. For any 
pitch, therefore, a separate cutter would be required corresponding 
to every numiber of teeth, to insure absolute accuracy. Practi- 
cally, however, this is not necessary. The change in the form 
of tooth outhne is much greater in a small gear, for any increase 
in the number of teeth, than in a large one. It is found that 
twenty-four cutters will cut all possible gears of the same pitch 
with sufficient practical accuracy. The range of these cutters 
is indicated in the following table, taken from Brown and Sharpe's 
*' Treatise on Gearing." 

These same principles of interchangeable sets of gears with 
cycloidal tooth outlines apply not only to small milled gears as 
above, but also to large cast gears with tooled or untooled tooth 
surfaces. 



TOOTHED IVHEELS OR GEARS 



293 











Table 


xx\q. 










itter A cuts 


12 


teeth 




Cutter M cuts 


27 to 


29 tee 


" B 




13 








N 




30 " 


2,2, " 


*' C 




14 













34 " 


37 " 


" D 




15 








P 




38'' 


42 " 


" E 




16 








Q 




43 " 


49 " 


" F 




17 








R 




50 " 


59 " 


" G 




18 








S 




60 " 


74 " 


" H 




19 








T 




75 " 


99 " 


" I 




20 








U 




100 " 


149 " 


" J 




21 


to 22 


teeth 




V 




150 " 


249 *' 


- K 




23 


" 24 


" 


< ( 


w 




250 " 


rack 


" L 




24 


" 26 


< < 




X 




rack 





193. Interchangeable Involute Gears. — In the involute system 
the second condition of interchangeability is that the angle between 
the common tangent to the base circles and the line of centers shall 
he constant. This may be shown as follows: Draw the line of 
centers, AB, Fig. 212. Through P, the assumed pitch point, 
draw CD, and let it be the constant common tangent to all base 




circles from which involute tooth curves are to be drawn. Draw 
any pair of pitch circles tangent at P, with their centers in the 
line AB. About these centers draw circles tangent to CD; these 
are base circles, and CD may represent a cord that winds from 
one upon the other. A point in this cord will generate simul- 
taneously involutes that will engage for the transmission of a con- 
stant velocity ratio. But this is true of any pair of circles that 
have their centers in AB, and are tangent to CD. Therefore, 



294 



MACHINE DESIGN. 



if the pitch is constant, any pair of gears that have the base circles 
tangent to the line CD will mesh together properly. As in the 
cycloidal gears, the involute tooth curves vary with a variation 
in the number of teeth, and, for absolute theoretical accuracy, 
there would be required for each pitch as many cutters as there 
are gears with different numbers of teeth. The variation is 
least at the pitch line, and increases with the distance from it. 
The involute teeth are usually used for the finer pitches and the 
cycloidal teeth for the coarser pitches ; and since the amount that 
the tooth surface extends beyond the pitch line increases with the 
pitch, it follows that the variation in form of tooth curves is 
greater in the coarse pitch cycloidal gears than in the fine pitch 
involute gears. For this reason, with involute gears, it is only 
necessar)^ to use eight cutters for each pitch. The range is shown 
in the following table, which is also taken from Brown and Sharpens 
"Treatise on Gearing": 

Table XXVII. 

No. I will cut wheels from 135 teeth to racks 

2 " 

3 " 

4 " 

5 " 



55 " 


'•' 134 


inclusive 


35 " 


" 54 




26 " 


" 34 




21 '' 


'' 25 




17 " 


" 20 




14 " 


'' 16 




12 '• 


" 13 





194. Laying Out Gear- teeth. — Exact and Approximate 
Methods. — ^There is ordinarily no reason why an exact method 
for laying out cycloidal or involute curves for tooth outlines should 
not be used, either for large gears or gear patterns, or in making 
drawings. It is required to lay out a cycloidal gear. The pitch, 
and diameters of pitch circle and describing circle are given. 

Draw the pitch circle on the drawing-paper, using a fine line. 
On a flat piece of tracing-cloth or thin, transparent celluloid draw 
a circle the size of the generating circle. Use a fine, clear line. 



TOOTHED IVHEELS OR GEARS. 295 

Place it over the drawn pitch circle so that it is tangent to the 
latter at P as shown in Fig. 213. AB is an arc of the pitch circle. 
Insert a needle point at P, and using it as a pivot swing the 
tracing-cloth in the direction of the arrow a very short distance, 
so that the generating circle cuts the pitch circle at a new point 
Q, as shown exaggeratedly in Fig. 214. Q should be taken very close 




Epicycloid 



Fig. 214. Fig. 215. Fig. 216. 

to P. Insert a needle point at Q, remove the one at P, and swing 
the cloth, about Q as a pivot, in the direction of the arrow until 
the two circles are tangent at Q. (See Fig. 215.) The point P 
of the tracing-cloth now lies off the pitch circle a short distance. 
With a needle point prick its present position through to the 
drawing-paper. Now with Q as a pivot rotate the tracing-cloth 
until the two circles intersect at a point R slightly beyond Q. 
Insert needle at R and remove the one at Q. Swing tracing- 
cloth about R until R becomes the point of tangency of the two 
circles and then prick the new position of P through to the drawing- 
paper. Taking points very near together and repeating the 
operation gives a close approximation to true rolling of the gen- 
erating circle on the pitch circle and therefore the path of the 
point P as marked on the drawing-paper by pricked points is an 
epicycloid and may be used for the face of the tooth. 

Next place the tracing-cloth on the inside of the pitch circle, 
as shown in Fig. 216, with the generating circle tangent to the 
pitch circle at the original point P. Using the method just 
described to prevent slipping, roll the generating circle, in the 
direction of the arrow, on the pitch circle and the path traced 
by P as marked by pricked points on the drawing-paper will be 
a hypocycloid for the flank of the tooth. 



295 



MACHINE DESIGN. 



The compound curve aPb, Fig. 217, has now been traced, 
which forms the basis of the completed tooth outhne. 

AB is an arc of the pitch circle whose center is at O. With 
O as center, swing in the addendum circle 'CD and the full 
depth circle EF, according to the proportions given in § 195. 
With a radius equal to j^j of the circular pitch draw the fillet 
cd tangent to EF and aPb. The completed tooth profile is the 
curve cdPe. 

Cut a wooden template to fit the tooth curve, and make it 
fast to a wooden arm free to rotate about O, making the edge of 
the template coincide with cdPe. It may now be swnmg succes- 
sively to the other pitch points, and the tooth outline may be drawn 
by the template edge. This gives one side of all of the teeth. 
The arm may now be turned over and the other sides of the 
teeth may be drawn similarly. 





It is required to lay out exact involute teeth. The pitch, 
pitch-circle diameter, and angle of the common tangent are 
given. — Draw the pitch circle AB, Fig. 218, and the line of 
centers 00\ Through the pitch point P draw CD, the common 
tangent to the base circles, m.aking the angle fi with the fine of 
centers. Draw the base circle EF about O tangent to CD. 

On a piece of flat tracing-cloth draw a fine, clear, straight line 
and lay the tracing-cloth over the drawing so that this line coin- 
cides with CD. Take a needle point and insert it at the point 



TOOTHED kVHEELS OR GEARS. 297 

of tangency Q. With another needle, mark the point P on the 
tracing-cloth. Now, employing a pair of needle points to pre- 
vent sHpping, roll the traced line on the base circle EF^ prick- 
ing the path, aPh, of the point P of the tracing-cloth through to 
the drawing-paper. This path is the involute of the base circle 
and is the basis of the involute tooth outline. To complete the 
latter proceed as follows: Draw the addendum circle GH and 
the full-depth circle JK. In general JK will lie inside of the 
base circle EF and it will be necessary to extend the tooth outline 
inward beyond a. About O swing a circle w^hose diameter equals 
one half the circular pitch and draw ac tangent to it. With a 
radius equal to jV o^ the circular pitch swing in the fillet de tan- 
gent to JK and ac. deaPj is the completed outline. If the 
gear has 20 teeth or less ac should be made a radial line. If 
EF lies inside of JK we draw the fillet tangent to JK and aP.* 

195. Gear Proportions. — The following formulas and table 
are given to assist in the practical proportioning of gears : 
Let D = pitch diameter; 
Di == outside diameter; 

Z>2= diameter of a circle through the bottom of spaces; 
P = circular pitch = space on the pitch circle occupied by 

a tooth and a space; 
p = diametral pitch =number of teeth per inch of pitch- 
circle diameter; 
N == number of teeth; 
t = thickness of tooth on pitch line; 
a = addendum; 

* Approximate tooth outlines may be drawn by the use of instruments, such 
as the WilHs odontograph, which locates the centers of approximate circular arcs; 
the templet odontograph, invented by Prof. S. W. Robinson; or by some geomet- 
rical or tabular method for the location of the centers of approximate circular 
arcs. For descriptions, see "Elements of Mechanism," Willis; "Kinematics," 
McCord; "Teeth of Gears," George B. Grant; "Treatise on Gearing," pub- 
lished by Brown and Sharpe. 



298 



MACHINE DESIGN. 



c = clearance; 

d = working depth of spaces; 

di =full depth of spaces. 



Then 



Ar+2 



N = 



p= 



t 
c= — 

10 



7Z 

p_ 

20 



i^l = 


p 


~ ; 


D2 = 


D~2(a+c) 


> 


Dn 
P ' 


p = 


Dk 

N 


D 


PN 

TV ' 


N- 


-Dp; 


P- 




D 


N 
~ P 


; Pp = 


--TZ\ 




; P 


7t 


t = 


P 

2 


TZ ■ 

— , no 
2p' 


backlash ; 



- — ; d = 2a\ di=2a + c: a=— inches. 

p20 p 



The following dimensions are given as a guide; they may be 
varied as conditions of design require: Width of face = about 3P; 
thickness of rim =1.25/; thickness of arms = 1.25^, no taper. 
The rim may be reinforced by a rib, as shown in Fig. 219. Diam- 
eter of hub=2Xdiameter of shaft. Length of hub = width of 
face + J''; width of arm at junction with hub=J circumference 
of the hub for six arms. Make arms taper about -J'' per foot on 
each side. 

Table XXVIII. 



Diametral 
Pitch. 


Circular Pitch. 


Thickness of 

Tooth on the 

Pitch Line. 


Diametral r,v^,-, 
Pitch. Circu 


lar Pitch. 


Thickness of 

Tooth on the 

Pitch Line. 


P 


P 


t 


P 


P 


t 


1 


6 
4 


283 
189 


3-141 
2.094 


3h 

4 


897 
785 


•449 
•393 


I 


3 
2 


141 
513 


I-57I 
1.256 


5 
6 


628 
523 


.314 
.262 


If 


2 


094 
795 


1.047 
.897 


7 
8 


449 
393 


.224 
.196 


2 
2J 




571 
396 
256 


.785 
.698 
.628 


9 

10 
12 


349 
314 
262 


.174 
•157 
•131 


2f 




142 


•571 


14 


224 


.112 


3 


1.047 


•523 









TOOTHED IVHEELS OR GEARS. 299 

196. Strength of Gear-teeth. — The maximum work trans- 
mitted by a shaft per unit time may usually be accurately estimated ; 
and, if the rate of rotation is known, the torsional moment may 
be found. Let O, Fig. 220, represent the axis of a shaft perpen- 
dicular to the paper. Let A = maximum work to be transmitted 
per minute; let iY= revolutions per minute; let Fr = torsional 
moment. Then F is the force factor of the work transmitted, 
and 27trN is the space factor of the work transmitted. Hence 

2F7trN = A , and Fr = torsional moment = — ^7. 

27: N 

If the work is to be transmitted to another shaft by means of 
a spur-gear whose radius is ri, then for equiHbrium Firi=Fr^ 

Fr 

and i^i = — . -Fi is the force at the pitch surface of the gear 

whose radius is ri, i.e., it is the force to be sustained by the gear- 
teeth. Hence, in general, the force sustained by the teeth of a gear 
equals the torsional moment divided by the pitch radius of the gear, 

WTien the maximum force to be sustained is known the teeth 
may be given proper proportions. The dimensions upon w^hich 
the tooth depends for strength are : Thickness of tooth = /, width 
of face of gear = 6, and depth of space between teeth =/. These 
all become known when the pitch is known, because / is fixed for 
any pitch, and / and b have values dictated by good practice. 
The value of b may be varied through quite a range to meet the 
requirements of any special case. 

In the design the tooth will be treated as a cantilever with 
a load apphed at its end. It is assumed that one tooth sustains 
the entire load; i.e., that there is contact only betw^een one pair 
of teeth. This would be nearly true for gears with low numbers 
of teeth; but in high-numbered gears the force would be distributed 
over several pairs, and hence they would have an excess of strength. 
It is also assumed that the load is uniformly distributed across 
the face of the tooth. This is a safe assumption if the width of 



?oo 



MACHINE DESIGN. 



face, h, does not exceed three times the circular pitch, i.e., t^P, and 
if the gears are well aligned and rigidly supported. All teeth of 
the same pitch have not the same form, as was explained in the 
discussion of interchangeable gears, and therefore they vary in 
strength. The fewer teeth the thinner they will be at their base 
and consequently the weaker they will be when acting as canti- 
levers. 






Fig. 219. 



Fig. 220. 



Fig 221. 



Mr. Wilfred Lewis * has drawn a number of figures on a 
large scale to represent ver\^ accurately the teeth cut by com^plete 
sets of cutters of the 15° involute, the 20° involute, and the cycloidal 
systems. In the latter he used a rolling circle having a diameter 
equal to the radius of the 12 -tooth pinion. The proportions 
of the teeth used in his investigation are slightly different from 
those given above which correspond to the Brown and Sharpe 
system, but no serious errors will result from applying the formulas 
derived by him. His reasoning was as follows (see Fig. 221): 

The greatest stress in the tooth occurs when the load is apphed 
at the end of the tooth as indicated by the arrow at a, its direction 
of action being normal to the profile at a. The component of 
this force, which is effective to produce rotation of the gear, equals 
F and is called the working force. 

This load is apphed at b and induces a transverse stress in 
the tooth. To determine where the tooth is weakest advantage 
is taken of the fact that any parabola in the axis be and tangent to 



* Proc. Phila. Eng. Club, 1893, and Amer. Mach., May 4 and June 22, 1893. 



TOOTHED H^HEELS OR GE/iRS. 



301 



hF incloses a beam uf uniform strength. Of all the parabolas 
that may thus be drawn one only will be tangent to the tooth form 
(as sho-wii by the dotted Une in the figure) and the weakest sec- 
tion of the tooth will be that through the points of tangency c 
and d. Having determined the weakest section in each case, Mr. 
Lewis developed the following general formulae from the data so 
obtained : 

For 15° involute and the cycloidal system, using a rolling 
circle whose diameter equals the radius of the 12-tooth pinion, 



F = jPh[o.i2^- 
For the 20° involute system 



0.684 

N 



} 



F.|Pi{o^^;t-°P) 



- * 



JP= working force in pounds; 

/=safe allowable unit stress in pounds per square inch; 
P= circular pitch in inches; 
& = width of face of gear in inches; 
N = number of teeth in the gear. 
Experimental data fixing the value of / for different materials 
and velocities are lacking. The following table is recommended 
by Mr. Lewis : 

Table XXIX. 



Velocity of pitch line 
in feet per minute . . 



/ for cast iron. 
/ for steel 



100 or 
less 


200 


300 


600 


900 


1200 


1800 


8000 


6000 


4800 


4000 


3000 


2400 


2000 


20000 


15000 


12000 


1 0000 


7500 


6000 


5000 



2400 



•(■JO ) 

4300 



* For the cycloidal system, using a rolling circle whose diameter equals the 

radius of the 15 -tooth pinion, 

0.67J 
F = fPb[o.io6- 



,(o.xo6-^^), 



(Trans A. S. M E., Vol. 18. p. 776.) 



302 MACHINE DESIGN. 

In these formulas, for a given gear the whole right side of 
the equation becomes known and the allowable value of F is 
readily determined. It is more difficult to apply the formulae 
where the force to be transmitted is given. In such a case the 
value of P is determined by trial. 

197. Problem. — Design a pair of 15° involute gears to transmit 
6 H.P. The distance between centers is 10 inches and the 
velocity ratio of the shafts is to be f . The pinion shaft makes 
150 revolutions per minute. 

The distance between centers being 10 inches and the velocity 
ration J, the radii will be to each other as 3:2 and their sum 
= 10 inches, hence the radius of the pinion will be 4 inches, while 
that of the gear will be 6 inches. If both gears are of the same 
material the teeth of the smaller will be the weaker. Com- 
putations will therefore be made for the pinion because the gear 
will be stronger and, consequently, safe. The pitch diameter 
of the pinion =8 inches, its velocity = 150 X^rX^V"" 3 14- 2 feet per 
minute. 

6 H.P. =6X33,000 = 198,000 ft. -lbs. per minute; 
_ 108000 



314.2 



= 632 lbs. 



Assuming cast-iron gears at 314.2 feet per minute, / may 
be taken =4800 lbs. The maximum value of h=2)P, but for 
first trial let 5 = 1 inch. For the 15° involute system, then, 

/ o.684\ 

632=4800X^X1(0.124— ^- ). 

^ ,^ r.D 25.13 

But iV=-^=^. 

Substituting this, transposing, and clearing of fractions, 
P2_4.56P + 4.84=o, 



TOOTHED IVHEELS OR GEARS. Z^Z 

from which P — 2.28 = ±c.6. 

Choosing the smaller value, 

P = i.68 inches. 

A pitch of 1.68 inches with a width of face of i inch is bad pro- 
portion. 

For our second trial let h = 2 inches, 

632 =4800X2 XP( 0.124- — 7^)' 

Solving this gives 

P = o.62 inch. 

This corresponds practically to a diametral pitch of 5, giving 
5X8=40 teeth for the pinion. The engaging gear will have 
5X12 =60 teeth. 

When the solution of the quadratic equation leads to an 
imaginary quantity it will be necessary to increase h. If the 
greatest allowable value of h still leaves the imaginary, then the 
value / must be increased either by using a stronger material 
or by cutting do^^Tl the factor of safety. 

Another way of solving the problem would have been to assume 
h in terms of P, thus h =KP (K being less than 3), thus giving 

Tn this substitute trial values for P until one is found Avhich 
sarisfics the equation. 

19S. Non-circular Wheels. — Only circular centrodes or pitch 
curves correspond to a constant velocity ratio; and by making 
the pitch curves of proper form, almost any variation in the velocity 



3^4 MACHINE DESIGN. 

ratio may be produced. Thus a gear whose pitch curve is an 
elHpse, rotating about one of its foci, may engage with another 
elHptical gear, and if the driver has a constant angular velocity 
the follower will have a continually varying angular velocity. 
If the follower is rigidly attached to the crank of a slider-crank 
chain, the slider will have a quick return motion. This is some- 
times used for shapers and slotting-machines. When more than 
one fluctuation of velocity per revolution is required, it may be 
obtained by means of "lobed gears"; i.e., gears in which the 
curvature of the pitch curve is several times reversed. If a 
describing circle be rolled on these non-circular pitch curves, the 
tooth outlines will vary in different parts; hence in order to cut 
such gears, many cutters would be required for each gear. Prac- 
tically this would be too expensive; and when such gears are 
used the pattern is accurately made, and the cast gears are used 
without "tooling" the tooth surfaces. 

199. Bevel-gears. — ^All transverse sections of spur-gears are 
the same, and their axes intersect at infinity. Spur-gears serve 
to transmit motion between parallel shafts. It is necessary also 
to transmit motion between shafts whose axes intersect. In this 
case the pitch cylinders become pitch cones; the teeth are formed 
^ upon these conical surfaces, the result- 
'^a ing gears being called bevel-gears. 
To illustrate, let a and h, Fig. 222, be 
the axes between which the motion is 
to be transmitted with a given velocity 

ratio. This ratio is equal to the ratio 
Fig. 222. ^ 

of the length of the line A to that of B. 
Draw a line CD parallel to a, at a distance from it equal to 
the length of the line A. Also draw the line CE parallel to h, 
at a distance from it equal to the length of the line B. Join 
the point of intersection of these lines to the pomt O, the 
intersection of the given axes. This locates the line OF, which 




TOOTHED IVHEELS OR GEARS. 305 

is the line of contact of two pitch cones which will roll together 

^ MC A , ., . 

to transmit the required velocity ratio. For vy-^ = — , and if it 

be supposed that there are frusta of cones so thin that they may 

be considered cylinders, their radii being equal to MC and NC, 

it follows that they would roll together, if shpping be prevented, 

to transmit the required velocity ratio. But all pairs of radii 

MC 
of these pitch cones have the same ratio, =T77^) and therefore 

any pair of frusta of the pitch cones may be used to roll together 
for the transmission of the required velocity ratio. To insure 
this result, slipping must be prevented, and hence teeth are 
formed upon the selected frusta of the pitch cones. The theo- 
retical determination of these may be explained as follows: 

200. ist. Cycloidal Teeth. — If a cone, A (Fig. 223), be rolled 
upon another cone, 5, their apexes coinciding, an element he of 
the cone A will generate a conical surface, and a spherical sec- 




tion of this surface, adh, is called a spherical epicycloid. Also 
if a cone, A (Fig. 224), roll on the inside of another cone, C, 
their apexes coinciding, an element he of A will generate a conical 
surface, a spherical section of which, hda, is called a spherical 
hypocycloid. If now the three cones, B^ C, and A, with apexes 
coinciding, roll together, always tangent to each other on one 
line, as the cylinders were in the case of spur-gears, there will 
be two conical surfaces generated by an element of A : one upon 
the cone B and another upon the cone C. These may be used 
for tooth surfaces to transmit the required constant velocity 



506 MACHINE DESIGN. 

ratio. Because, since the line of contact of the cones is the axo * 
of the relative motion of the cones, it follows that a plane normal 
to the motion of the describing element of the generating cone 
at any time will pass through this axo. And also, since the 
describing element is always the line of contact between the gen- 
erated tooth surfaces, the normal plane to the line of contact of 
the tooth surfaces always passes through the axo, and the condi- 
tion of rotation with a constant velocity ratio is fulfilled. 

201. 2d. Involute Teeth. — If two pitch cones are in contact 
along an element, a plane may be passed through this element 
making an angle (say 75°) with the plane of the axes of the cones. 
Tangent to this plane there may be two cones whose axes coin- 
cide with the axes of the pitch cones. If a plane is supposed to 
wind off from one base cone upon the other, the line of tangency 
of the plane with one cone will leave the cone and advance in the 
plane toward the other cone, and will generate simultaneously 
upon the pitch cones' conical surfaces, and spherical sections of 
these surfaces will be spherical involutes. These surfaces may 
be used for tooth surfaces, and will transmit the required con- 
stant velocity ratio, because the tangent plane is the constant 
normal to the tooth surfaces at their line of contact, and this 
plane passes through the axo of the pitch cones. 

202. Determination of Bevel-gear Teeth. — To determine the 
tooth surfaces with perfect accuracy, it would be necessary to 
draw the required curves on a spherical surface, and then to 
join all points of these curves to the point of intersection of the 
axes of the pitch cones. Practically this would be impossible, 
and an approximate method is used. 

If the frusta of pitch cones be given, B and C, Fig. 225, then 
points in the base circles of the cones, as Z, M, and K^ will move 
always in the surface of a sphere whose projection is the circle 

* An axo is an instantaneous axis, of which a centre is a projection. 



TOOTHED IVHEELS OR GEARS. 



307 



LA KM. Properly, the tooth curves should be laid out on the 
surface of this sphere and joined to the center of the sphere to 
generate the tooth surfaces. Draw cones LGM and MHK tan- 
gent to the sphere on circles represented in projection by lines 
LM and MK. They are called the ''back cones." If now 
tooth curves are dravm on these cones, with the base circle 
of the cones as pitch circles, they will approximate the tooth 
curves that should be dra\\Ti on the spherical surface. But a 
cone may be cut along one of its elements and rolled out, or de- 
veloped, upon a plane. Let MDH be a part of the cone MHK, 
developed, and let MNG be a part of the cone MGL, developed. 




Fig. 225. 




The circular arcs MD and MN may be used just as pitch circles 
are in the case of spur-gears, and the teeth may be laid out in 
exactly the same way, the curves being either cycloidal or in- 
volute, as required. Then the developed cones may be wrapped 
back and the curves drawn may serve as directrices for the 
tooth surfaces, all of whose elements converge to the center of 
the sphere of motion. 

203. Cutting Bevel-gear Teeth. — The teeth of spur-gears may 
be cut by means of milling-cutters, because all transverse sections 
are alike, but with bevel-gears the conditions are different. The 
tooth surfaces are conical surfaces, and therefore the curvature 
varies constantly from one end of the tooth to the other. Also the 



30 8 MACHINE DESIGN. 

thickness of the tooth and the width of space vary constantly 
from one end to the other. But the curvature and thickness 
of a milhng-cutter cannot vary, and therefore a milling-cutter 
cannot cut an accurate bevel-gear. Small bevel-gears are, 
however, cut with milling-cutters with sufficient accuracy for 
practical purposes. The cutter is made as thick as the narrowest 
part of the space between the teeth, and its curvature is made 
that of the middle of the tooth. Two ciits are made for each 
space. Let Fig. 226 represent a section of the cutter. For the 
first cut it is set relatively to the gear blank, so that the pitch 
point a of the cutter travels toward the apex of the pitch cone, 
and for the second cut so that the pitch point h travels toward 
the apex of the pitch cone. This method gives an approximation 
to the required form. Gears cut in this manner usually need 
to be filed slightly before they work satisfactorily. Bevel-gears 
with absolutely correct tooth surfaces may be made by planing. 
Suppose a planer in which the tool point travels always in some 
line through the apex of the pitch cone. Then suppose that as it 
is slowly fed down the tooth surface, it is guided along the required 
tooth curve by means of a templet. From what has preceded 
it will be clear that the tooth so fonned will be correct. Planers 
embodying these principles have been designed and constructed 
by Mr. Corliss of Providence, and Mr. Gleason of Rochester, 
with the most satisfactory results. 

204. Design of Bevel-gears. — Given energy to be transmitted, 
rate of rotation of one shaft, velocity ratio, and angle between 
axes; to design a pair of bevel-gears. Locate the intersection 
of axes, O, Fig. 227. Draw the axes OA and OB, making the 
required angle with each other. Locate OC, the line of tan gen cy 
of the pitch cones, by the method given on p. 304. Any pair of 
frusta of the pitch cones may be selected upon which to form 
the teeth. Special conditions of the problem usually dictate this 
selection approximately. Suppose that the inner linait of the 



TOOTHED IVHEELS OR GEARS. 309 

teeth may be conveniently at D. Then niake DP, the width 
of face, ^DO-^2. Or, if P is located by some limiting condi- 
tion, lay off PD=PO^^. In either case the limits of the teeth 
are defined tentatively. Now from the energy and the number 
of revolutions of one shaft (either shaft may be used) the moment 
of torsion may be found. The mean force at the pitch surface 
= this torsional moment -^ the mean radius of the gear; i.e., 
the radius of the point M, Fig. 227, midway between P and D. 
The pitch corresponding to this force may be found. 

In order to compute it consider the teeth of the pinion {i.e., 
the smaller gear), as they will be the weaker. Having found 
the force, F, w^hich is to be transmitted we determine the pitch 
required to carry F by a spur-gear, whose pitch radius =ilfiV 
and whose width of face, b, equals PD. (The radius MN is 
used to govern the shape of the tooth and not MR, because the 
teeth are laid off on the developed back cones and not on the 
pitch circles, as has been explained. The circle whose radius 
is MN is sometimes called the formative circle in order to dis- 
tinguish it from the pitch circle.) 

The pitch of such a spur-gear would be the mean pitch of the 
bevel-gearsc But the pitch of bevel-gears is measured at the 
large end, and diametral pitch varies inversely as the distance 
from O. In this case the distances of M and P from O are to 
each other as 5 is to 6. Hence the value of diametral pitch found 
X|-=the diametral pitch of the bevel-gear. If this value does 
not correspond with any of the usual values of diametral pitch, 
the next smaller value may be used. This would result in a 
slightly increased factor of safety. If the diametral pitch thus 
found, multiplied by the diameter 2PQ, does not give an integer 
for the number of teeth, the point P may be moved outward 
along the line OC, until the number of teeth becomes an integer. 
This also would result in slight increase of the factor of safety. The 
point P is thus finally located, the corrected width of face =P0 -^3, 



3IO 



MACHINE DESIGN. 



and the pitch is known. The drawing of the gears may be com- 
pleted as follows: Draw AB perpendicular to PO. With A and 
B as centers, draw the arcs PE and PF. Use these as pitch arcs, 
and draw the outlines of two or three teeth upon them, with 




Fig. 227. 

cycloidal or involute curves as required. These will serve to show 
the form of tooth outlines. From P each way along the Hne AB 
lay off the addendum and the clearance. From the four points 
thus located draw lines toward O terminating in the Hne DG. 
The tops of teeth and the bottoms of spaces are thus defined. 
Lay off upon AB below the bottoms of the spaces a space HJ 
about equal to f the thickness of the tooth on the pitch circle. 
This gives a ring of metal to support the teeth. From / draw 
JK toward O. The web L should have about the same thickness 
as the ring has at K. Join this web to a properly proportioned 
hub as shown. The plan and elevation of each gear may now 
be drawn by the ordinary methods of projection. Use large 
fillets. 



TOOTHED IV HEELS OR GEARS. 



311 



205, Skew Bevel-gears. — When axes which are parallel are 
to be connected by gear-wheels the basic form of the wheels is the 
cylinder. When intersecting axes are to be so connected the basic 
form is the cone or cone frustum. It is sometimes necessary to 
communicate motion between axes that are neither parallel nor 
intersecting. If the parallel axes are turned out of parallelism, 
or if intersecting axes are moved into different planes, so that they 
no longer intersect, the pitch surfaces become hyperboloids of revo- 
lution in contact with each other along a straight line, which is the 
generatrix of the pitch surfaces. These hyperboloids of revolu- 




FiG. 227a. 



tion rotated simultaneously about their respective axes, circum- 
ferential sHppage at their line of contact being prevented, will 
transmit motion with a constant velocity ratio. There is, however, 
necessarily a shppage of the elements of the surfaces upon each 
other parallel to themselves. Teeth may be formed on these pitch 
surfaces, and they may be used for the transmission of motion 
between shafts that are not parallel nor in the same plane. Fig. 
227a shows, in plan view, a pair of such hyperboloids of revolu- 
tion. Disk portions of these, cut anywhere except at the gorge, 
are approximately conical frusta and are the basic form of skew 
bevel-gears. The difficulties of construction and the additional 
friction due to slippage along the elements make them undesir- 
able in practice, and there is seldom a place where they cannot be 




312 MACHINE DESIGN. 

replaced by some other form of connection. It is evident from 
the figure, for instance, that disk portions taken at the gorge of 
the hyperboloids of revolution are approximately cylinders, which 
are the basic forms of ordinary spiral gears. 

A very complete discussion of skew bevel-gears may be found 
in Reuleaux's "Constructor." 

206. Spiral Gearing.* — If line contact is not essential 
there is much wider range of choice of gears to connect shafts 
which are neither parallel nor intersecting. A and B, Fig. 228^ 
are axes of rotation in different planes, both 
planes being parallel to the paper. Let EF and 
GH be cylinders on these axes, tangent to each 
other at the point S. Any line may now be 
drawn, in the plane which is tangent to the 
two cylinders, through 5 either between A and 
B or coinciding with either of them. This Hne, say DS, 
may be taken as the common tangent to hehcal or screw 
lines drawn on the cylinders EF and GH; or hehcal surfaces 
may be formed on both cylinders, DS being their common tan- 
gent at S. Spiral gears are thus produced. Each one is a por- 
tion of a many threaded screw. The contact in these gears is 
point contact; in practice the point of contact becomes a very 
Hmited area. 

For the sake of simplicity the special case of spiral gears with 
axes at 90° will be considered first. The term helix angle is here 
taken (as in the treatment of all other screws) as meaning the 
angle of the mean helix with a plane which is perpendicular to 
the axis of rotation. Throughout the discussion the subscript i 
is used in reference to the driver and the subscript 2 in reference 
to the follower. 



* See further "Worm and Spiral Gearing," by F. A. Halsey. Van Nost rand 
Science Series 



TOOTHED fVHEELS OR GEARS. 



2>n 



Let Fig. 228a represent the plan view of a pair of spiral gears 
with axes at 90°. In this special case it will be noted that the 
axis of the follower lies in the plane perpendicular to the axis of 
the driver and, therefore, that the helix angle of the driver is the 
angle ABC, made by the teeth of the driver with the axis of the 
follower. Call this helix angle of the driver «!. 

Similarly, the hehx angle of the follower equals the angle be- 
tween the teeth of the follower and the axis of the driver. Call 
the heHx angle of the follower a2. 





-o^ 



Fig. 228a. 



Fig. 228b. 



Let Fig. 228Z) represent the development of both pitch cyl- 
inders in the tangent plane. It will be noted that the same line 
is normal to the teeth of each gear and that their pitches measured 
on the normal must be equal. The distance occupied by a tooth 
and space on this normal is called the normal pitch. 

If now the driver (i) be moved in the direction AB, the 
follower (2) will be forced to move in the direction CD. For 
a movement of the driver equal to ah, the follower must move a 
distance ch. This establishes the fact that 



circumferential vel. of follower 
circumferential vel. of driver 



distance moved by follower ch 
distance moved bv driver ah° 



But — = tan 0:1; 
ah 



circumferential vel. of follower 
circumferential vel. of driver 



tan ai. (i) 



314 MACHINE DESIGN. 

Let C= distance between centers of gears; 

Z)i = pitch diameter of driver; 

jD2 = pitch diameter of follower; 

iVi = number of teeth of driver; 

iV2 = number of teeth of follower; 
r.p.m.i = revolutions per minute of driver; 
r.p.m. 2 = revolutions per minute of follower. 

The following equations may then be written: 

circumferential velocity of follower 7:D2 r.p.m.2 

= , , (2) 

circumferential velocity of driver nDi r.p.m. 1 

Combining (i) and (2), 

;rZ>2 r.p.m.2 



ttDi r.p.m. 1 . 

D2 r.p.m.i 
Di r.p.m.2 



= tan ai; 



tanai . (3) 



p-vT T-. • / \ 11 • r.p.m.2 Di 

[Note. — Equation (3) may also be written = — - tan ai. 

r.p.m.i D2 

But -^— ^ — - is the angular velocity ratio. Hence in spiral gears 
r.p.m.i 

the angular velocity ratio depends upon two factors: first, the 

inverse of the ratio of pitch diameters; second, the tangent of 

the helix angle of the driver. 

In ordinary spur-gears, it will be remembered, the angular 
velocities are inversely as the pitch diameters, or the pitch 
diameters are inversely as the velocities. In spiral gears (axes 
at 90°) this is only true when tan ai = i, hence when the helix 
angle is 45°. 

Whereas with spur-gears, for a given distance between centers 
and a given velocity ratio, the pitch circles are at once deter- 
mined since there can only be a single pair to fit the conditions; 



TOOTHED IVHEELS OR GEARS, 315 

with spiral gears an indefinite number of values may be used 

for D\ and D2. It is only necessary to keep =the given 

2 

center distance, and — - tan ai = the required velocity ratio. As 
D2 

Di and D2 are varied it is only necessary to vary tan ai accord- 
ingly; or, if tan ai is varied — - must be varied accordingly. 

D2 

For gears with axes at 90°, if 0:1 = 45°, 0:2 = 45° ^^^o, and the 
diameters will be inversely as the angular velocities, i.e., in- 
versely as the revolutions per minute. For all other values of ai 
this does not hold.] 

Since C= distance between centers 

Di + Z>2=2C; (4) 

/. D2=2C-Di. (5) 

Substituting in (3), 

2C— Z)i r.p.m.i 

tan aif 



Di r.p.m.2 

^ -r. ^ /r.p.m.i \ 

2C-Di = Di( -^ — i tan 0:1 , 

\r.p.m.2 / 

^ ^ / r.p.m.i \ 

2C=Di i + — tan 0:1 , 

\ r.p.m.2 / 



.p.m.2 

Di= — ■ (6) 

r.p.m.i 

i + — ^^ tan ai 

r.p.m.2 

Equations (6) and (5) will give us all the possible solutions for 

r.p.m.i 



any given values of C and 



r.p.m.2 



3i6 



MACHINE DESIGN, 



The particular solution which will best suit a given case is 
determined by practical considerations. 

First, it must be remembered that spiral gears have screw 
action and hence have highest efficiencies for helix angles whose 
value is in the neighborhood of 45°. For satisfactory operation 
it must never be allowed to be less than 15° or over 75°. 

Second, it must be remembered that as ai approaches 45° 
the ratio of the diameters becomes the inverse of the velocity 



Normal Ilolix 



B-^ 




Fig. 228c. 



ratio. For a very great velocity ratio this may make one of 
the gears too small and the other too large for practical con- 
struction and operation. 

Third, the gears must be such as can be cut with stock 
cutters in any universal milling machines. For this reason 
their normal pitch must have some standard value and the lead 
of the tooth helix must also be a value which can be attained 
with the milling machine. 

The following discussion w^ill deal with this third condition. 
See Fig. 228c, A, which shows a spiral gear extended axially 
to a length sufficient to include a complete tooth-helix. 



TOOTHED IVHEELS OR GEARS. 317 

The normal helix is a helix at right angles to the tooth helix. 
Its entire length is that length which will include (i.e., cut across) 
all the teeth of the gear exactly once. 

Let Zi = length of normal helix of driver; 
1.2 = length of normal helix of follower; 
P = normal pitch of each. 
From Fig. 228c, jB, which shows a developed pair of gears, 
it can be seen that 

Li = DiS\nai', (7) 

Z2 = Z>2 cos ai; (8) 



From equation (3), 



rr^'"""" ^9^ 



r.p.m.2 Di 

=— -tanai, 

r.p.m.i D2 

Li r.p.m.2 , , 

.*. -^ = (10) 

L2 r.p.m.i 

That is, the normal helices are inversely as the number of 
revolutions per minute. Since 

Li = length of normal hehx of driver, 
P=^ normal pitch, 
and iVi = number of teeth of driver, 

Zi^PiVi. 

Similarly, L2 == PN2, 

Li Ni r.p.m.2 , , 

.. -;r-=zrzr = (ll) 

L2 N2 r.p.m.i 

That is, the numbers of teeth are inversely as the numbers of 
revolutions. (Just as in spur-gears.) 





Pp- 


= n; 


,; 


P-- 


7Z 






-P; 


_•, -. 


'7' 


•" 


•■ P= 


ii ■ 




L2 

N2 


-P; 


.-. 


P = 




> 



31S MACHINE DESIGN. 

From (11), iVi=?^^5^^iV2. 

r.p.m.i 

iVi must be a whole number, as must also N2\ =— — -Nijy 

\ r.p.m.2 / 

for a fractional tooth is impossible. Hence a normal pitch 
must be selected which will divide both Li and L2 perfectly. 

Also this normal pitch must be a stock value. Let p be 
the diametral pitch corresponding to the normal pitch P. Then 



But 
Also 



and p must correspond to some standard diametral pitch. 

The values of Zi and Z2 as determined by equations (7) 

and (8) may not be such as will give tabular values for p. It 

then becomes necessary to take the nearest standard value for 

p to that obtained from p = —r~ and to substitute it in this 

Li 

equation, thus deriving a corrected value for Li. Similarly for 
L2. This is based upon the assumption that Ni and N2 are 
given. If p is given, A^i and N2 must be computed, the nearest 
corresponding whole numbers selected, and Li and L2 corrected 
accordingly. 

But changing Li and L2 involves also changing Di, D2, 
aij and 0:2 as can be seen by reference to (7) and (8), 

Li = 7zDi sin ai, (7) 

L2 = 7iD2Cos ai, (8) 

{ — nD2 sin 0:2). 



TOOTHED IVHEELS OR GEARS. 319 

If ai be not altered, it is evident that Di and D2 will be 
altered proportionately with Zi and L2 and this will change 

the value of the center distance C, since C= . 

2 

If the center distance may be altered to this new value the 
solution is complete. The size of the gear blank of the driver 

2 
would be Di-i — ; helix angle, a 1; number of teeth, ATi ; normal 

P 
(diametral) pitch, p. For the follower the values would be D2 -f • 

2 

-; 0:2 ( = 90°-a:i); N2, and p. 
P 

In most cases, however, it will be impossible to change the 
value of C, and the values of Di, D2, cci, and 0:2 must all be 
changed to keep the corrected values of Li and L2. 

From (7) and (8), 

n ^^ n ^^ ' 

V\ = : , 1J2 = ; 

7t sm a\ 71 cos ai 

also, Z>i + D2 = 2C (4) 



U ^ L2 ^ 

\ =2C. 



Ti sm ai Tz cos ai 



Divide by — -. , 

Ti sm a\ 



L2 2C . . . 

i+— tanai = — - sm ai (12) 



Using corrected values of Li and L2, try different values for 
ai until we get an identity. This is the correct value of «!. 
Use this value of ai and the correct values of Li and L2 in 

Di = — : and D2 



71 sm ai 7z cos ai 

and solve for corrected values of Di and Z>2. 



320 



MACHINE DESIGN. 



These corrected values of Di, D2, oli, and a2 (=9o°-ai), 
together with Ni, N2, and p, already obtained, fully determine 
the gears. 

There remain two points of practical importance to be deter- 
mined: first, the " pitch of the tooth helix;" second, the particular 
cutter of the determined pitch which should be used. 

I. By pitch of the tooth helix is meant the axial length 
corresponding to one complete turn of the tooth helix about 
the pitch cyhnder. In ordinary screws this is termed the " lead," 
and as " pitch " is used for so many different purposes we will 
use the term '' lead." 

Referring to Fig. 228c, B, it is clear that 

leadi 



:D, 



tan «!, 



.*. leadi = 7rZ>i tan ai. 



lead; 



tan a2==cot ai; 



kD2 

.'. lead2 = 7:Z)2 cot ai. 

From these leads the gear settings of the milling machine 
are determined. (See Halsey's " Worm and Spiral Gears " 
for table of Brown and Sharpe settings.) 

2. In ordinary spur-gears the cutter to be used for any gear 

is directly determined by the 
number of teeth of the gear. 
This is not the case with spiral 




The method for determining 
the cutter is based upon the 
following reasoning, due to Prof. Le Conte. 

Reference is to Fig. 2 28 J. a = helix angle as before. 

The figure shows the material of the pitch cylinder extended 



TOOTHED IVHEELS OR GEARS. 321 

either side of the gear; ahc is the tooth helix; np represents a 
plane normal to the tooth helix at b. This normal plane will 
cut an ellipse from the pitch cylinder. The minor axis will 
= D, the diameter of the pitch cylinder. The major axis is 

determined bv the relation — -, r- = sina; .*, major axis 

major axis 

D 

sin a 

If we cut a spiral gear in the same way as we cut this pitch 
cylinder, selecting the point h midway between two teeth, the 
form of the space on the normal plane will be the true normal 
shape. It will therefore be the true shape of the cutter to be 
used. 

Now the curvature of the normal section of the gear at the 
point indicated is, of course, the curvature of the ellipse at the 
extremity of the minor axis. And the cutter to be used would 
be the cutter for a pitch circle having this curvature. Such a 
circle (i.e., one whose radius equals the radius of curvature of 
the ellipse at the extremity of its minor axis) is called an " oscu- 
lating circle." 

Let ,0 = radius of osculating circle; 

a = half of major axis of elhpse 



^ = half of minor axis of ellipse 



2 sm a 
D 



_, a^ 4 sin- ex D 

Then, ,= ^ =__ = __ (,3) 



2 



Let A^o = number of teeth of normal pitch P on the osculat- 
ing circle, 

.-. 7Vo=^ (14) 



322 MACHINE DESIGN. 

Combining (i) and (2), 



No = ^^^~^ (15) 



Let AT = actual number of teeth on spiral gear of diameter D, 

P' = actual circular pitch of spiral gear. 
Then, 

^=p7 (16) 

P 
It will also be seen that — > = sin a, 



P AT ^^ sin a r. ^P 

sm a P sm a 



Substituting value of D of (17) in (15), 



NP N 
sm-^ aP sm^a 



For the driver, then 



For the follower, 



iVo = — ^ (19) 



AT ^2 iV2 . , 

^0 = ^^= ^ — (20) 

sm'^Q;2 cos'^ai 



Equations (19) and (20) give the number of teeth whose 
corresponding cutters should be used. 

This completes the solution for spiral gears with axes at 90°. 

The following problem gives the full application of the fore- 
going method. The computation of spiral gears which will run 
together properly calls for strictly accurate numerical work and 
the use of logarithmic tables is recommended. 



TOOTHED IV HEELS OR GEARS. 323 

206a. Problem. — Design a pair of spiral gears for the follow- 
ing conditions: 

Axes at 90°. C = 3.375''. 
r.p.m.2 I revolutions per minute of follower 
r.p.m.i 2 revolutions per minute of driver ' 
Z>i = pitch diameter of driver; 
Z>2 = pitch diameter of follower; 
a 1 = helix angle of driver; 
0:2 = helix angle of follower (=90°— a 1); 
iVi = number of teeth of driver =10; 
iV2 = number of teeth of follower = 20; 
Zi = normal hehx length of driver; 
Z<2 = normal helix length of follower. 
It is further assumed that, for reasons of construction, it is 
desired to have the two gears as nearly equal in size as possible. 

First Solution. 
Let Di = D2, and allow C to change in value. 
Z>2 r.p.m.i 



Di r.p.m.2 



tanai (3) 



2 
i = - tan ai; 



,*. tan q:i = 4; 
•. 0:1 = 26° 34', .-. a2 = 6f 26'. 
Trial Li=--Di sin ai 

=--X3.375X.4472 
= -X 1.509. 
Trial L2 = r,D2 cos ai 

= -X3.375X.8944 
= -X3.oi8. 



324 MACHINE DESIGN. 

But ^ = -7 — and Ni has been assumed = 10; 

.*. p = = 6.63. 

^ 71X1.509 ^ 

The nearest standard diametral pitch to this is 7. Selecting 
^ = 7, with iVi = io, 

Actual, or corrected, Li = NiP = , 

P 

.. Correct — = — = — =1.420. 

Actual, or corrected, L2 = N2P=N2 -, 

P 

. ^ L2 N2 20 

.. Correct — = — = — =2.8^8. 
r p 7 

_ _ correct Zi , , 1.420 

Correct Di = -. , 7) =—^-^ = 3.195^ 

7: sm «! .4472 

2 
Driver gear-blank diameter = Z)i + - = 3.i95''4-.286'' = 3.48i'^ 

P 

_ _ correct L2 ,^, 2.8 s8 

Correct D2 = -, (8) =-^ = 3.195-. 

n cos ai .8944 

2 
Follower gear-blank diameter = 1)2 H — = 3.i95''4-.286'' = 3.48i''. 

P 

To select the cutter of the 7 diametral pitch set: 



iVi _ 10 
sin^ai .4472- 



Driver No=-r-^ — = = 111.8, 



calling for B. & S. involute cutter No. 2. 



N2 _ 20 
sin3a2 .8944' 



FoUower iVo=:^5-= ^1773=27-95; 



calling for B. & S. involute cutter No. 4. 



TOOTHED J^'HEELS OR GEARS. 325 

To determine the lead of tooth helix, in order to select the 
corresponding gear set of the milling machine: 

Driver lead = 7:i)i tan ai 

= 7rX3.i95X.5 
==5.018-, 
calling for gears 86, 48, 28, and 100 in B. & S. milHng machine. 

Follower lead = 71^)2 cot ai 
= 71X3.195X2 
= 20.075-, 

calling for gears 86, 24, 56, and 100 in B. & S. milling machine. 
Summary for modified distance between centers: 



Driver. 


Follower. 


Pitch diameter, 2)1 = 3.195- 


^2 = 3-195" 


Gear-blank diameter = 3.481- 


Blank diameter = 3.481- 


Number of teeth, Ni = 10 


Ar2=20 


Helix angle, cti = 26° 34' 


0:2=63° 26' 


Diametral pitch, />=7 


p=l 


Cutter, involute, No. 2 


Cutter No. 4 


■Lead of tooth heHx = 5.018- 


Lead =20.075- 


Gears, 86, 48, 28, 100 


Gears, 86, 24, 56, 100 


^ D1 + D2 

2 


= 3-195". 


Second Solution. 



Taking the same data and assuming that the center dis- 
tance remains fixed at 3.375'', it is still desired to have Z>i 
and D2 as nearly equal as possible. 

If iVi = io, N2 = 20, and ^ = 7 it is fixed that 
£1 = 71X1.429, 
and Z2 = 71X2.858. 



326 MACHINE DESIGN. 

^ . , s L2 2C . 

From equation (12) i +— tan «! = — - sin ai, 



6.75 . 

.'. i + 2tanai = sm ai, 

1.429 

or, 1 + 2 tan ai = 4.724 sin ai. 

The next step is to substitute trial values of ai until a value 
is found which will make the two sides of the equation equal. 
If the right-hand member comes out greater than the left the 
trial value of ai is too large; if the left-hand member comes 
out greater than the right the trial value of a 1 is too small. 

Starting with a trial value of 0:1 = 26° 34', from the first 
solution, it is found to be too large. A few trials lead to the 
value of 23° 5' for ai which gives 

I -^ 2 X .42619 = 4.724 X .39207, 
.'. 1.852 = 1.852. 

Therefore the correct 0:1 = 23° 5', and 

correct 0:2 = 90° — 01 = 66° 55' 

_ ^ correct Li 1.429 ^ „ 

Correct Di = —. -=:^L^ = 3.645''. 

TT sm oi (correct) .3921 

^ ^ correct L2 2.858 

Correct Z>2= } r= ^=3.106''. 

7T COS ai (correct) .91993 

2 
Driver gear-blank diameter =Di + -=3. 645'' -{-.286 = 3. 931''. 

P 

2 
Follower gear-blank diameter = Z)2 + -- = 3 . 106'' -f- .286'' = 3.392''. 

P 

Driver No = '^-- = ~o= 165.9, 

sm-^oi .3921"* 

calling for B. & S. involute cutter No. i. 



TOOTHED IVHEELS OR GEARS. 327 

N2 20 

Follower Nq= — -— = 5 = 25.1^ 

cos%i .9199"* 

calling for B. & S. involute cutter No. 5. 

Driver tooth-helix lead = 7rZ)i tan ai 

= ;rX3.645X.4262 
= 4.88'^ 
calling for B. & S. gear-set, 48, 64, 56, 86. 

Follower tooth-helix lead = 7cD2 cot ai 

= 7rX 3. 106X2.3463 
= 22.9'', 

calling for B. & S. gear-set, 72, 44, 56, 40. 

Summary for fixed distance between centers: 

Driver. Follower. 

Pitch diameter, Z>i = 3.645'' Z)2=3.io6" 

Gear-blank diameter = 3.931" Blank diameter = 3.392'' 
Number of teeth, A^i = 10 ^^2 = 20 

HeHx angle ^1 = 23° 5' 0:2 = 66° 55' 

Diametral pitch p = '] ^=7 

Cutter, involute No. i Cutter No. 5 

Lead of tooth hehx =4.88'' Lead =22.9'' 

Gears, 48, 64, 56, 86 Gears, 72, 44, 56, 40 

c= — ^ — =3.375 . 

206b. Spiral Gears with Axes at any Angle, /?. — Fig. 228^ 
shows a plan view of such a pair of gears, and also a view of the 
gears developed in the tangent plane. 

From the latter it is evident that a motion ba of the driver 
in its direction of rotation must induce a motion he of the 



328. 



MACHINE DESIGN, 



follower in its direction of rotation. This establishes the fact 
that 

circumferential velocity of follower be 
circumferential velocity of driver ba' 




LEFT HAND SPIRALS 



,^'-^■'-1 




Fig. 228e. 



Consider the triangle abc. Angle cab=ai, angle ac6=a2, 

be sin ai 
' ' ba sin ^2 * 

circumferential velocity of follower nD2 r.p.m.2 
circumferential velocity of driver nDi r.p.m.i' 

7rjD2 r.p.m.2 sinai 



TtDi r.p.m.i sina2' 

D2 r.p.m.i sin ai 
Di r.p.m.2 sin 0:2' 



(I) 



TOOTHED IVHEELS OR GEARS. 329 

r.p.m.2 D\ sin ai , , 

or, — = 77-^ (2) 

' r.p.m.i Do sm «£ 

Di + D2=^2C, 

:. D2=2C-Di (3) 

Substitute in (i) 

2C — D1 r.p.m.i sin ai 



Di r.p.m.2 sin 0:2' 

\r.p.m.2 sm a'2/ 

^ ^ / r.p.m.i sin aA 

2C=Z)i 1 + -^^ ^, -), 

\ r.p.m.2 sm 0:2/ 



.p. 

^ 2C 

.-. Z)i = r . ..... (4 

r.p.m.i sm ai 

IH : 

r.p.m.2 sm 0:2 

Because ai+^ + a2=iSo°, sin a2 = sin 05+«i) and (4) may 
also be written 

A= ^- — (5) 

r.p.m.i sm ai 
r.p.m.2 sin (/? + a:i) 

[Note. — Equations (4) or (5) and (3) give us all possible 

solutions for anv value of C and -^-^ — '—, just as when the axes 

r.p.m.2 

were at 90°. In fact (5) reduces to the form used in that case 

when /5 = 9o°, for 

2C .2C 



Dv 



r.p.m.i sm ai r.p.m.i sm ai 

1 + 



r.p.m.2 sin (go^ + ai) r.p.m.2 cos ai 

2C 



r.p.m.i 

i + — '^ tanai. 

r.p.m.2 



330 MACHINE DESIGN. 

It can be shown (Reuleaux's '' Constructor ") that the 
sliding velocity (along ac) of the teeth upon each other is least 
when ai = a2', .'. whenever possible these values should be 
given «! and 0:2. That is, ai = a2 = i{iSo°—(^). 

We have already seen that when /?=90°, ai = a2 = 45° is the 
most efficient angle of helix. 

It must be borne in mind that these values of ai, 0:2, Di, 
and D2 may give us impractical values for normal pitch and 
that, in consequence, the values may have to be modified to get 
a normial helix length which will give an even number of teeth 
of stock size.] 

As in spiral gears with axes at 90° we have 

Li = 7rDi sin ai, (6) 

L2=7tD2 sin a2 (7) 

But Li = NiP = Ni-. 

P 

Li Ni ( Ni CSC ai^ 

.*. lJi = — : = ——. . . . [p) 

Tt sm «! p sm ai 



/ Ni CSC aA 
\ P / 



N\ = pDi sin ai; (q) 

^ N2 / N / iV2CSCa2\ 

D2= — -. , (10) = 

p sin a2 \ P ■' 

N2'=pD2 sin «2; (j-i) 

^ r. . r. ATi CSCa:i + iV2CSCa:2 , . 

2C = Di-\-D2 = . . . . (12) 



The foregoing equations may be used to get the practical 
solution. 

206c. Problem. — The following problem will illustrate the 
method. Compute a pair of spiral gears, 

Shaft angle, /?=4o°. 



TOOTHED IVHEELS OR GEARS. 33 1 

Exact center distance = 3'' (not to be changed), 

r.p.m. 1 = 400, 

r.p.m.2 = 3oo, 
p=io. 

Solution. 

a:i + a2=i8o°-40°-i40°, 

Try ai=-a2= =70, 

From equation (4) , 

_ 2C 2C 6 6 „ 

Z>i = -. --= = — = =2.57"; 

r.p.m.i sm ai r.p.m.i 4 2.333 

r.p.m.2 sin 0:2 r.p.m. 2 3 

Z)2 = 2C-Z>i = 6- 2.57 = 3.43''. 
From (9), Ni^pDi sin ai, 

A/'i = ioX2.57X.94 = 24.i. 
From (10), N2=pD2 sin 0:2, 

iV2= 10X3.43 X. 94 = 32. 13- 

■^oo 
But Ni and N2 must be whole numbers in ratio of - — , or 

400 

24 and 32, respectively. 

Use these values of Ni and iV2. 
Substitute in (12) 

iVi CSC ai +A7'icsca2 
O = , 

2p 

24X 1.064 + 32 X 1.064 



332 MACHINE DESIGN. 

It is evident that new values of ai and a 2 will have to be tried 
to make this equation an identity. It is necessary to take trial 
values of ai and a2, remembering that the sums of ai and az 
must always be 180° — /? (=140° in this case). 

Tryai = 74°, 0:2 = 66°. Substitute in (12) 

24 X 1.0403 X 32 X 1.0946 



3= ixi3 ^-9997. 



which will answer. 



17 /Qx ^ r» ATicscai 24X1.0403 ,, 

From (8), correct Z)i = = ^ = 2.497". 

P 10 

^ , . ^ A/'2CSca:2 32X1.0946 

From (10), correct D2 = =- — =3.503". 

2 
Driver gear-blank diameter = Z)iH — = 2. 497'' + .2'' = 2.697"* 

P 

2 
Follower gear-blank diameter = Z>2 + - = 3'S^7'' + .2" = 3.703". 

P 



Lead of tooth-helix and cutter to be used are found as in spiral 
gears with axes at 90°. 

From the nature of hyperboloidal wheels, two solutions are 
always possible, depending upon whether /? or its supplement 
be taken in determining the line of contact of the hyperboloids. 
In this problem it is evidently just as proper to consider the 
shaft angle to be 140° as 40°. See Fig. 228/. 

Here 



^2- 


ai) = i8o°, 




a2- 


-ai = i8o°- 


-A 




= 180°- 


-140°, 




= 40°. 





TOOTHED IVHEELS OR GEARS. 
Trya2=7o°, 0:1 = 30°. 

Di- '^. = 3.51", 



1 + 



r.p.m.i sin ai 
r.p.m.2 sin a2 



1)2=2.49' 



y 



.-'^'"J 



di 



s'l 





\ 


^ 
1 


; 




-^ 


1 


^ ^ Axis o« 


SPIRALS 
DRIVER LEFT HAND 


/ 


-V * Follower 


FOLLOWER RIGHT HAND, 


/ 


^ 


/ 




^ --. ; ^-^ 


X 






-i-A^'^"^^ 




Fig. 2 


28y 





333 



c= 



Trial iVi = 17.55, A^2=23.4 (from (9) and (10)), 
Correct, iVi = 18, A^2 = 24. 
A^i CSC ai + iY2 CSC a2 ^ 18X2 + 24X1.064 



• o 



20 



3.0768, 



showing necessity of modifying a\ and a^. 
Try 71° 15' and 31° 15' for ai and 0:2, 

18X1.9276 4- 24X1.056 

.*. J?= = S.002. 

^ 20 ^ 

Hence these values of a\ and 0:2 will answer. 

^ ^ A^l CSC «! 

Correct D\ = = 3.47 , 



Correct 2^2 = 



P 

N2 CSC «2 

1 =2.53''. 



334 MACHINE DESIGN, 

[It is to be noted that in the two cases the spirals have differ- 
ent relations. In the first case where the spirals are both of 
the same hand, /?+ 0:1 + 0:2=180°, i.e., 

ai + a2=i8o°-/?. 

In the second case, where the spirals are of opposite hands, 

/?+ (0:2-0:1) = 180°, 
i.e., 0:2 — «i = i8o°— /?. 

The topic of direction of spirals and direction of rotation 
is well treated in the American Machinist^ Oct. 11, 1906.] 

207. Worm-gearing. — When the angle between the shafts is 
made equal to 90°, and one gear has only one, two, three, or four 
threads, it becomes a special case of spiral gearing known as 
Worm- gearing. In this special case the gear with a few threads 
is called the worm, while the other gear, which is still a many- 
threaded screw, is called the worm-wheel. If a section of a worm 
and worm-wheel be made on a plane passing through the axis 
of the worm., and normal to the axis of the worm-wheel, the form 
of the teeth will be the same as that of a rack and pinion; in 
fact the worm, if moved parallel to its axis, would transmit rotary 
motion to the worm-wheel. From the consideration of racks 
and pinions it follows that if the involute system is used, the sides 
of the worm-teeth will be straight lines. This simplifies the cutting 
of the worm, because a tool may be used capable of being sharpened 
without special methods. If the worm-wheel were only a thin 
plate the teeth would be formed Hke those of a spur-gear of the 
same pitch and diameter. But since the worm-wheel must have 
greater thickness, and since all other sections parallel to that 
through the axis of the worm, as CD and AB, Fig. 229, show a 
different form and location of tooth, it is necessary to make 
the teeth of the worm-wheel different from those of a spur-gear, 
if there is to be contact between the worm and worm-wheel any- 
where except in the plane EF, Fig. 229. This would seem to 
involve great difijculty, but it is accomplished in practice as 



TOOTHED IVHEELS OR GEARS, 



335 



follows : A duplicate of the worm is made of tool steel, and '' flutes " 
are cut in it parallel to the axis, thus making it into a cutter, 
which is tempered. It is then mounted in a frame in the same 
relation to the worm-wheel that the worm is to have when they 
are finished and in position for working. The distance between 
centers, however, is somewhat greater, and is capable of being 
gradually reduced. Both are then rotated with the required 
velocity ratio by means of gearing properly arranged, and the 
cutter or ''hob " is fed against the worm-wheel till the distance 

between centers becomes the required value. The teeth of the 

.c 





Fig. 229 Fig. 230. 

worm-wheel are "roughed out " before they are "bobbed." By 
the above method the worm is made to cut its own worm- 
wheel.* 

Fig. 230 represents the half section of a worm. If it is a 
single worm the thread A, in going once around, comes to B\ 
twice around to C, and so on. If it is a double worm the thread 
A, in going once around, comes to C, while there is an inter- 
mediate thread, B. It follows that if the single worm turns 
through one revolution it will push one tooth of the worm-wheel 
with which it engages past the line of centers; while the doiible 
worm will push two teeth of the worm-wheel past the lint' of 
centers. The single worm, therefore, must make as many reyolu- 
tions as there are teeth in the worm-wheel, in order to cause 
one revolution of the worm-wheel; while for the same result 
the double worm only needs to make half as many revolutions. 



* This subject is fully treated in Unwin's "Elements of Machine Design,'' 
and in Brown and Sharpe's "Treatise on Gearing." 



336 MACHINE DESIGN. 

The ratio of the angular velocity of a single worm to that of 

n 
the worm-wheel with which it engages is=— , in which n equals 

the number of teeth in the worm-wheel. For the double worm 
this ratio is — . 

2 

Worm-gearing is particularly well adapted for use where it 
is necessary to get a high velocity ratio in limited space. 

The pitch of a worm is measured parallel to the axis of rota- 
tion. The pitch of a single worm is P, Fig. 230. It is equal 
to the circular pitch of the worm-wheel The pitch of a double 
worm is Pi =2P = 2 Xcircular pitch of the worm-wheel. 

208. Design of Worm-gears. — All spiral gears are forms of 
screw transmission and the formulae for efficiency, etc., developed 
under c, s:c. 98, in the chapter on Screws, apply to them directly. 

Three points are to be carefully considered in the design of 
worms and wheels: 

1. Speed of rubbing. This is the velocity in feet per minute 
of a point on the pitch line of the worm. The best efficiencies 
are obtained when this is about 200 feet per minute. When 
it exceeds 300 feet there is increasing danger of cutting, and 
the pressure on the teeth must be correspondingly reduced. 
At high speeds (say 1000 feet) only very light pressure can be 
sustained without abrasion. 

2. Pressure on teeth. This depends on the speed and on the 
angle of helix. 

3. Angle of helix. From the formula for screw efficiency we 
have seen that this should be made as great as convenient provided 
it does not exceed 45°. Practical conditions make it impossible 
to use the highest values, but 20° gives very excellent results. It 
should never be less than 15° for fair efficiency. 

Oil-bath lubrication should be used wherever possible; faiHng 
this, a heavy mixture of graphite and oil has been found satis- 



TOOTHED IV HEELS OR GEARS. 



337 



factory. The following table based on Professor Stribeck's 
experiments * applies to a 20° angle of helix and oil-bath lubri- 
cation, using a hardened-steel worm and phosphor-bronze wheel. 

Table XXX. 



Rubbing velocity in teet per minute 

Allowable pressure for maximum efficiency 
in pounds 



300 
?7oo 



400 



500 600 
12^0 1000 



This may be taken as a guide. When the angle is greater 
than 20° the values of the pressure may be slightly increased. 
When the angle is less than 20° they should be rapidly diminished ; 
thus for 10° use only one half the value given. 

Since worms and wheels are simply spiral gears in which 
one of the gears has a very few teeth, all of the general formulae, 
relating to Di, D2, ai, a'2, p, etc., developed in the preceding 
sections, are directly applicable in their design. However, as 
worms are usually cut in a lathe (like screw threads) and worm- 
wheels hobbed, it is the axial pitch of the worm, equal to the 
circumferential pitch of the wheel, which is of importance rather 
than the normal pitch. This " lead " must have a value obtain- 
ble with the screw-cutting gearing of the lathe. It is therefore 
practically more convenient in the design of worms and wheels 
to follow the method illustrated by the following problem: 

209. Problem. — Two shafts about 10 inches apart and at a 
right angle with one another are to have a velocity ratio of 20 to i. 
The worm-shaft makes 300 revolutions per minute. 

Since the velocity ratio is 20 to i, the wheel will have to have 
20, 40, or 60 teeth, depending upon whether the worm is single-, 
double-, or triple-threaded. 

If the shafts are 10 inches apart the greatest allowable pitch 
radius of the wheel will not be far from 8 inches ; 50 inches may 
be taken as a trial pitch circumference of the wheel. 



* Zeitschrift d. Vereins deutscher Ingenieure. 



S3^ MACHINE DESIGN. 

With a single-thread worm this will give a circular pitch of 
|-g- = 2j inches. With a double thread the circular pitch would 
be 1^ =ii inches. 

In any case the rise of the pitch helix of the worm will be 
2j inches for one revolution. 

This value must always be such that the thread may be cut in 
an ordinary lathe. 

If it is required that the helix angle, a, be 20°, then the 
pitch circumference of the worm must be such that 

tan 20° = ^-5 ' 

°* pitch circumference of worm ; 

2.5 
.*. pitch circumference of worm = — — =6.87 inches. 
^ 0.364 

Pitch diameter of worm =— — =2.2 inches. 
Pitch diameter of wheel = — = i "^ . 88 inches. 

15.88 + 2.2 

Actual distance between shafts = =9-04 inches* 



The question now arises whether 2.2 inches is a great enough 
pitch diameter for the worm. If the thread is single the pitch 
= 2.5 inches and the corresponding dedendum=o.92 inch. 

Twice this dedendum = i.84 inches, which subtracted from 
2.2 inches would only leave a central solid core of 0.36 inch diam- 
eter for the worm. It is obvious without computation that this 
would not sustain the torsional moment. If the double thread 
were used the central core would have a diameter of 1.28 inches. 

For each revolution of the worm the length of the path of 
the point of contact or the distance rubbed over equals the helix 
length on the pitch line. This is the hypothenuse of a right- 



TOOTHED IVHEELS OR GEARS. 339 

angle triangle whose base =6.87 inches and whose altitude = 
2.5 inches, or 7.3 inches. 

At 300 revolutions per minute the distance rubbed through in 

feet per minute = 300 X -^- = 182 feet. At 182 feet the allowable 

pressure, W, between the teeth may equal 3500 lbs., assuming 
bath lubrication, a steel worm, and a bronze wheel. This is the 
pressure apphed at the circumference of the woim-wheel in the 
direction of the axis of the worm. The total work done on the 
worm.-wheel in foot-pounds per minute will equal W multiplied 
by the pitch velocity of the wheel in feet per minute. 

This wheel makes Vo^- =^ 15 revolutions per minute and its 
pitch circumference =-f-| feet, hence its pitch velocity = 15 Xft 
= 62.5 feet per minute. 

3500X62.5 =218,750 ft.-lbs. =6.63 H.P. 

This same amount of energy is transmitted through the worm. 
The twisting moment on the shaft =i^r, where r equals the pitch 
radius of the worm. F=energy transmitted -^ the velocity of 
the point of application of the force. 

^ 2187S0 „ 

300 x^ 

P'r=i274Xi.i==i40i in.-lbs. 
To resist this there is a circular section whose strength is 
represented by/. . 



2 

ri = radius of core of worm = 0.64 inch; 
/s = unit stress in outer fiber; 

. r Fr 1401 

. . /* = — , - — = 3400 lbs. 



340 MACHINE DESIGN, 

This is a safe value for steel. Therefore the double-threaded 
worm will be used and the wheel will have 40 teeth of ij inches 
circular pitch. 

For the strength of the worm-wheel teeth Professor Stribeck 
gives : 

l^ = 35oP6 for cast-iron wheels; 
W -=^^oPh for bronze wheels; 
IF = pressure in direction of worm axis in pounds; 
P -= circular pitch in inches ; 
^=arc EF, Fig. 231, in inches. 

It will be impossible to make h as great as this would call for 
in the problem if IF = 3500 and P = i J inches. 

Had the distance between the shafts been fixed at 10 inches 
the hehx angle could not have been assumed but must have been 
calculated. 

The pitch radius of worm would have been 



Pitch circumference = 12.69 inches. 

2.5 
Tangent of hehx angle = — 7— =0.1955. 

/. a; = ii° + . 

With the center distance fixed at 10 inches, the helix angle 
need not necessarily be as low as 11°; provided, of course, that 
the axial pitch of the worm may be changed to some other value 
greater than 2.5''. As a check on the final results the general 
formulae of the preceding sections may be applied to the values 
obtained by the method here followed. 

When the worm and worm-wheel are determined, a working 
drawing may be made as follows: Draw AB, Fig. 231, the axis 
of the worm-wheel, and locate O, the projection of the axis of 



TOOTHED IVHEELS OR GEARS. 



341 



the worm, and P, the pitch-point. With O as a center draw the 
pitch, full depth, and addendum circles, G, H, and K\ also 
the arcs CD and EF, bounding the tops of the teeth and the 
bottoms of the spaces of 
the worm-wheel. Make the 
angle /3 = 90°. Below EF lay 
off a proper thickness of 
metal to support the teeth 
and join this by the web LM 
to the hub N. The tooth 
outlines in the other sectional 
view are drawn exactly as 
for an involute rack and 
pinion. Full views might 
be drawn, but they involve 
difhculties of construction, 
and do not give any addi- 
tional information to the 
workman. The drawing 
should contain a clear state- 1-231 

ment of the size and form of the worm tooth, the lead, whether the 
worm is single, double, triple, or quadruple threaded, the number 
of teeth of the wheel, and its helix angle, in addition to all ordinary 
dimensions. 

210. Compound Spur-gear Chains. — Spur-gear chains may 
be compound, i.e.^ they may contain links which carry more than 
two elements. Thus in Fig. 232 for links a and d each carry 
three elements. In the latter case the teeth of d must be counted 
as two elements, because by means of them d is paired with both 
h and c. In the case of the three-link spur-gear chain, Fig. 197, 
the wheels h and c, meshed with each other, and a point in the 
pitch circle of c moved with the same linear velocity as a point in the 
pitch circle of h, but in the opposite sense. In Fig. 232 points in 
all the pitch circles have the same linear velocity, since the motion 
is equivalent to roUing together of the pitch circles without slip- 
ping; but c and h now rotate in the same direction. Hence it 




342 



MACHINE DESIGN. 



is seen that the introduction of the wheel d has reversed the 
direction of rotation, without changing the velocity ratio. The 
size of the wheel, d, which is called an "idler," has no effect upon 
the motion of c and b. It simply receives upon its pitch circle 
a certain linear velocity from c, and transmits it unchanged 
to b. Hence the insertion of any number of idlers does not 
affect the velocity ratio of c to b, but each added idler reverses 
the direction of the motion. Thus, with an odd number of 
idlers, c and b will rotate in the same direction; and with an 
even number of idlers c and b will rotate in opposite directions. 

If parallel lines be drawn through the centers of rotation of 
a pair of gears, and if distances be laid off from the centers on 
these lines inversely proportional to the angular velocities of the 
gears, then a line joining the points so determined will cut the 
line of centers in a point which is the centro of the gears. In Fig. 
232, since the rotation is in the same direction, the lines have to 




Fig. 232. Fig. 233. 

be laid off on the same side of the line of centers. The pitch 
radii are inversely proportional to the angular velocities of the 
gears, and hence it is only necessary to draw a tangent to the pitch 
circles of b and c, and the intersection of this line with the line of 
centers is the centro, be, of c and b. The centrodes of c and b are 
Ci and bi, circles through the point be, about the centers of e and b. 
Obviously this four-link mechanism may be replaced by a three- 
link mechanism, in which Ci is an annular wheel meshing with a 
pinion bi. The four Hnk mechanism is more compact, however, 
and usually more convenient in practice. 

The other principal form of spur-gear chain is shown in 



TOOTHED IVHEELS OR GEARS. 343 

Fig. 233. The wheel d has two sets of teeth of different pitch 
diameter, one pairing with c and the other with h. The point 
hd now has a different Hnear velocity from cd, greater or less in 
proportion to the ratio of the radii of those points. The angular 
velocity ratio may be obtained as follov/s: 



also, 



angular velocity d _ C . . . cd 
angular velocity c D . . . cd' 

angular velocity h _ D . . .hd 
angular velocity d B . . . bd' 



. angular velocity b C . . . cdxD . . . bd 

' ^ ^ °' angular velocity c JD . . . cdxB . . . bd' 

The numerator of the last term consists of the product of the 
radii of the "followers," and the denominator consists of the 
product of the radii of the "drivers." The diameters or numbers 
of teeth could be substituted for the radii. 

In general, the angular velocity of the first driver is to the 
angular velocity of the last follower as the product of the number 
of teeth of the followers is to the product of the number of teeth 
of the drivers. This applies equally well to compound spur-gear 
trains that have more than three axes.* Therefore, in any spur- 
gear chain the velocity ratio equals the product of the number of 
teeth in the followers divided by the product of the number of 
teeth in the drivers. The direction of rotation is reversed if the 
number of intermediate axes is even, and is not reversed if the 
number is odd. If the train includes annular gears their axes 
would be omitted from the number, because annular gears do 
not reverse the direction of rotation. 

A common modification of the chain of Fig. 233 is shown in 
Fig. 233 A. Here the axis of the gear c is made to coincide with 

* Epicyclic trains excepted. 



344 MACHINE DESIGN. 

the axis of &, and the mechanism is known as a reverted gear train. 
Probably the best known application of this mechanism is that of 
the backgearmg of the ordinary engine lathe. The velocity ratio 
of c and h is, of course, not altered by having their axes coincide, 
and it is equally evident that one of them only may be keyed to 
the shaft while the other is free to rotate on it. 





Fig. 233 a. Fig. 233 B. 

2 20 a. Epicyclic Gearing. — In the gear trains of the preceding 

sections, the velocity ratios have been studied with reference to 

a fixed member to which each gear is attached by a turning pair. 

Fig. 233 B illustrates such a simple chain of three links, a, b, and 

c. Considering a as fixed it is evident that, if c makes m turns 

per minute relatively to a, causing b to make n turns per minute 

n 
relatively to a, for one turn of c relatively to a, b will make — turns 

m 

ft 
relatively to a. The ratio — is called the velocitv ratio, and is 

m 

designated by r. 

If, now, one of the gears, c, be considered as the fixed link, 

and it is desired to examine the action of the mechanism when 

a is swung about ac as center, it is evident that a different 

mechanism is obtained. See §8 and § 12. The action can be 

explained under the general laws laid down in these sections 

but can be understood more readily by reference to Fig. 233 C- 

Such mechanisms are known as epicyclic gear trains, because 

points in the one gear describe epicycloidal curves relatively to 

the other gear. The name has no connection with the form of 

the gear teeth which may belong to the cycloidal, involute, or 

any other system. 



TOOTHED IVHEELS OR GEARS. 345 

^ Let it be supposed that the three links can be rigidly locked 
together and while so held are given a complete turn about the 
axis ac, in a clockwise direction. Owing to this, h will make 
one turn in a clockwise direction about its own axis ah. In 
position I the arrow is seen to be horizontal, and to the left of ab, 
at 2 it is vertical and above ah^ at 3 horizontal and to the right; 
at 4 vertical and below, and at i, when the turn about ac ha? 
been completed, it is once more hoiizontal and to the left of ah. 



riG. 233 c. 

The arrow on h has, therefore, made a complete turn about ah 
as axis, and if one line of the rigid body h has made such a turn 
the whole body h has done so. But in swinging the locked 
mechanism about ac, the link c has been given a complete revo- 
lution in a clockwise direction. This is contrary to the original 
assumption that c be the fixed link, i.e., remain at rest. If, 
now, the mechanism be unlocked and c be given a complete 
revolution in a counter-clockwise direction while a is held sta- 
tionary, the result will be the same as though c had not been 
allowed to move at all. But this counter-clockwise revolution 
of c will cause h to- have a further clockwise rotation about its 

axis of — = r turns. The total number of turns which h makes 
m 

about its axis while a makes one turn about ac will, therefore, 
equal i -f r. 



346 



MACHINE DESIGN. 



Had an idler been placed between h and c, the result would 
have been to cause h to be given r turns in a counter-clockwise 
or negative direction, when c was brought back to its original 
position and, consequently, h would make i— r revolutions about 
its axis for one revolution of a about ac. This can be seen 
clearly in Fig. 233 D, where h and c are purposely made the 




4 — \— , 



Fig. 233 r. 

same size so that r =1 and, hence, i—r=o. In other words, 
in this special case the gear h does not rotate about its axis 
at all; its motion, as can be seen from positions i, 2, 3, and 4, 
being merely translation, as the arrow on h remains always parallel 
to its origmal position. 

A second intermediate gear, or idler, would again reverse 
the direction of &'s motion, making the revolutions of h =i-\-r. 

The general law may be stated as follows: — "The number of 
revolutions made by the last wheel of an epicyclic train for each 
revolution of the arm is equal to the one plus the velocity ratio 



TOOTHED IV HEELS OR GEARS. 347 

of the train if the number of axes in the train be even, and one 
minus the velocity ratio of the train if the number of the axes be 
odd. In the former case the wheel turns in the same sense as 
the arm ; in the latter in the opposite sense, unless the ratio r 
Is less than unity." (Kennedy — Mechanics of Machinery.) 

The same holds if there are no annular gears in the train or 
if there are an even number of them. If, however, there be one 
or any other odd number of annular gears in the train, the effect 
will be to transpose the plus and minus as well as the sense of 
rotation. 

If the first wheel of any epicyclic train has its axis fixed, but 
has itself a motion of rotation about this axis so that, for example, 
it makes k revolutions for each revolution of the arm, then the 
last wheel of the train will make i±r±kr revolutions instead 
of J±r. The sign of r is determined as before but the sign of 
kr is plus, if the rotation of the first wheel causes the last wheel 
to rotate in the same sense as the arm, and minus, if the rotation 
of the first wheel causes the last wheel to rotate in a sense opposite 
that of the arm. 

The only case which requires special attention for fear of 




Fig. 233 E. 

incorrectly determining the number of axes is where the gear 
train of Fig. 233, which has three axes, is given the reverted form 
shown in Fig. 233 E, which apparently has but two axes. For 
proper analysis it is necessary to consider the reverted train 
the same as the original form, i.e., a double axis is counted as 
two single ones in computing the number of axes in the train. 



348 MACHINE DESIGN. 

Problem. — Find the number of revolutions c will make about 
its axis for each revolution of the arm a\ d being considered as 
the fixed link. 

d has 1 01 teeth and meshes with h which has loo teeth, h' is 
keyed to same shaft as &, has 99 teeth, and meshes with Cy which 
has 100 teeth. If this were an ordinary reverted gear train with 
a as fixed link, then, remembering that the angular velocity of 
the last follower is to the angular velocity of the first driver as 
the product of the number of teeth of the drivers is to the product 
of the number of teeth of the followers, for one turn of d, c would 

make ^ = turns in the same sense. This is r, the 

100 X 100 looco 

velocity ratio of the train. Considering the train as an epicyc- 
lic one with d as fixed link, there are three axes and no annular 
gears and the rule would be that for one turn of a in a clock- 
wise direction c would make 1 — r turns about its axis in 

, 1 OQQQ I 

the same sense, equal to i — = • 

1 0000 1 0000 



CHAPTER XVIII. 

SPRINGS. 

211. Springs Defined. — Usually machine members are required 
to sustain the applied forces without appreciable yielding and are 
designed accordingly; but certain machine members are useful 
because of considerable yielding. They are generally called springs. 

212. Illustrations. — (a) The spring of a safety-valve on a steam- 
boiler holds the valve down until the steam-pressure reaches the 
maximum allowable value; then it yields and allows steam to 
escape until the pressure is reduced, when it closes the valve. 

(b) The springs upon which a locomotive-engine is supported 
prevent the transmission of the full effect of the shocks, due to 
running, to the working parts of the engine, thereby reducing the 
resulting stresses. Car-springs in a similar manner protect 
passengers and freight. 

(c) "Bumper" springs reduce stresses in cars and their con- 
tents due to axial shocks. 

(d) The springs in certain steam-engine governors yield under 
the increased centrifugal force of the governor weights, due to 
increased rotative speeds, and allow the adjustment of the valve- 
gear to the changes of effort and load. 

(e) Heavy reciprocating parts are often brought to rest without 
shock and are then helped to start on their return travel by the 
expanding spring. 

(/) A power-hammer strikes a "cushioned blow" because of the 

action of a spring. This spring may be of steel, rubber, or steam. 

(g) Belt connections are really yielding members and tend 

349 



3.SO 



MACHINE DESIGN. 



to reduce shocks transmitted through them; while gears (except 
rawhide or "hard-fiber ") yield almost imperceptibly and trans- 
mit shocks almost unchanged. 

(k) Long bolts may become springs for the reduction of 
stress due to shock. 

{i) Springs may serve for the storing of energy which is 
given out slowly to actuate light-running mechanisms, like clocks. 

213. Cantilever Springs. — Many springs are simple cantilevers 
with end loads. (See Fig. 234.) 



^ ^_ 



Fig. 234. 



— ^- — * 



Fig. 235. 



The rectangular spring of constant width h and height (or 
thickness) h, with a load F applied at a distance / from the sup- 
port gives, from the laws of beams, 



F = ^ and i=;g^. 



/ is the unit stress in the outer fiber in pounds per square inch, 
all forces being expressed in pounds and all dimensions in inches ; 
J is the total deflection in inches due to the application of F; 
E is the modulus of elasticity of the material used. 

For a flat spring of uniform breadth b, rectangular cross- 
section, top surface flat and lower surface a parabola in outline, 
such as is shown in Fig. 235, 



F = 



6/ 



and J = 



6FP 
Ebh^' 



SPRINGS. 



351 



The same equations hold approximately for the cantilever 
spring sho-wn in Fig. 236. They also hold for the triangular 
spring of constant depth h shown in Fig. 237. 







jo 




*- 7 -* 




1 




+ 2 


rf^ 




T-.1 


t 


F 


F 



^7t 




Fig. 236. 



Fig. 237. 



In all of these cases obviously the yielding varies inversely 
as h^, and the strength directly as h^; hence, if h be increased 
to obtain required strength, the yielding will be decreased as the 
cube of h while the strength is increased only as the square of h. 
Much of the requisite yielding is therefore sacrificed if the strength 
is obtained by increasing h. 

Inspection of the same equations shows that increasing the 
breadth b to obtain the required strength decreases the deflection 
in the same proportion. In springs, 
therefore, where yielding is to be kept 
large, it is better to gain requisite 
strength by varying b; while in a beam 
h should be as large as possible because 
here deflection is to be reduced to the 
smallest value. If the spring is to be of 
tool steel, hardened and tempered, thin 
material is better suited to the operation 
of hardening. 

As b is increased with a constant 
small value of h, it may become too great for the available 
space. This difficulty is overcome as shown by reference to 
Fig. 238. 

Suppose ABC is a triangular spring designed for certain con- 




FiG. 238. 



352 MACHINE DESIGN. 

ditions of load and yielding. AB is an inconvenient width. 
Divide AB into equal parts, say six. Conceive the portion 
GFHBi cut off and placed in the position BiLKiEi, and simi- 
larly conceive that EDKAi occupies the position AiMKiEi^ 
and the two parts are rigidly joined along the line KiEi. Also 
conceive the portion ADE moved to AiDiEi, and BFG moved 
to BiDiEi, and that they are rigidly joined along the line DiEi. 

The amount of material is unchanged. The bending force is 
applied in nearly the same way to the portions whose position is 
changed. The leaf spring is therefore practically equivalent 
to the triangular spring from which it is made. 

214. Springs for Axial Loads. — Many springs are subjected 
to axial loads; they are usually helical in form as shown in 
Fig. 239. 





Fig. 239. 

F may act to stretch or compress the spring. 
Consider the cross-section of the rod to be circular. 
Let F=load in pounds; 

c/= diameter of rod in inches; 
a = mean radius of coil in inches; 
AT = number of coils; 
/=developed length of spring in inches =2;ra AT; 
/«= allowable unit shearing stress in outer fiber ip pounds 

per square inch; 
£g= modulus of shearing elasticity = f£ ; 
J = extension or compression in inches. 
Then the following equations may be developed : 






'^^ .=3-^^, and N='^ 



SPRINGS. 353 

In a helical spring for an axial load using a rectangular cross- 
section of wire the axial height of wire = h and radial breadth = b, 
the equations become 

It will take one and a half times as much material to make a spring 
of this type as it would to make a round -wire helical spring of 
equal strength. 

215. Springs for Torsional Movements. — Many springs come 




Fig. 240. 



imder class {i) as mentioned above. The general case is shown 
in Fig. 240. The spring is a spiral of fiat wire with an axial 
height b and radial depth h. 

F = turning force in pounds ; 

R = lever-arm in inches ; 

/ = unit stress in outer fiber, pounds per square inch; 

/= developed length of spring; 

■& = angular deflection ; 

J = distance in inches moved through by the point of appli- 
cation of F. 

Then i^=W and J=R^= „,,„ . 

6R Ebh^ 

For further information the reader is referred to Reuleaux's 
"Constructor*' and Trans. A. S. M. E., Vols. V and XVI. 



CHAPTER XIX. 

MACHINE SUPPORTS. 

2 1 6. General Laws for Machine Supports. — The single-box 
pillar support is best and simplest for machines whose size and 
form admit of its use. When a support is a single continuous 
member, its design should be governed by the following principles : 

I. The. amount of material in the cross-section is determined 
by the intensity of the load. If vibrations are also to be sus- 
tained, the amount of material must be increased for this purpose. 

II. The vertical center hne of the support should coincide 
with the vertical line through the center of gravity of the part 
supported. 

HI. The vertical outlines of the support should taper slightly 
and uniformly on all sides. If they were parallel they would 
appear nearer together at the bottom. 

IV. The external dimensions of the support must be such 
that the machine has the appearance of being in stable equihbrium. 
The outline of all heavy members of the machine supported must 
be either carried without break to the foundation, or if they 
overhang, must be joined to the support by means of paraboHc 
outHnes, or by the straight lines of the brace form. 

217. Illustration. — In Fig. 241 the first three principles may 
be fulfilled, but there is an appearance of instabihty. It is 
because the outline of the "housing" overhangs. It should be 
carried to the foundation without break in the continuity of the 
metal, as in Fig. 242. 

354 



MACHINE SUPPORTS. 



355 



2i8. Divided Supports. — When the support is divided up into 
several parts, modification of these principles becomes necessary, 
as the divisions require separate treatment. This question may 




Fig. 241. Fig. 242. 

be illustrated by lathe supports. In Fig. 243 are shown three 
forms of support for a lathe, seen from the end. For stability 
the base needs to be broader than the bed. In A the width of 
base necessary is determined and the outlines are straight lines. 
The unnecessary material is cut away on the inside, leaving legs 
which are compression members of correct form. The cross - 
brace is left to check any tendency to buckle. For convenience 
to the workmen it is desirable to narrow this support somewhat 




Fig. 243. 

Without narrowing the base. The cross -brace converts the single 
compression member into two compression members. It is allow- 
able to give these different angles with the vertical. This is done 
in B and the straight Hnes are blended into each other by a curve. 
C shows a common incorrect form of lathe support, the compres- 
sion members from the cross -brace downward being curved. 
There is no reason for this curved form. It is less capable of 



356 



MACHINE DESIGN. 



bearing its compressive load than if it were straight, and is no 
more stable than the form B, the width of base being the same. 




'TT^L 



Fig. 244. 



Consider the lathe supports from the front. Four forms are 
shov^Ti in Fig. 244. If there were any force tending to move the 
bed of the lathe endwise the forms B and. C would be allowable. 
But there is no force of this kind, and the correct form is the 
one shown in T). Carrying the foot out as in ^, 5, and C in- 
creases the distance between supports (the bed being a beam 
with end supports and the load between); this increases the de- 
flection and the fiber stress due to the load. This increase in 
stress is probably not of any serious importance, but the prin- 
ciple should be regarded or the appearance of the machine will 
not be right. If the supports were joined by a cross-member, 




Fig. 245. 




as in Fig. 245, they would be virtually converted into a single 
support, and should then taper from all sides. 

219. Three-point Support. — If a machine be supported on 
a single -box pillar, change in the form of the foundation cannot 
induce stress in the machine frame tending to change its form. 
If, however, the machine is supported on four or more legs the 



M/ICHINE SUPPORTS. 357 

foundation might sink away from one or more of them and leave 
a part unsupported. This might cause torsional or flexure stress 
in some part of the machine, which might change its form and 
interfere with the accuracy of its action. 

But if the machine is supported on three points this cannot 
occur, because if the foundation should sink under any one of 
the supports the support would follow and the machine would 
still rest on three points. When it is possible, therefore, a ma- 
chine which cannnot be carried on a single pillar should be sup- 
ported on three points. Many machines are too large for three- 
point support, and the resource is to make the bed, or part sup- 
ported, of box section and so rigid that even if some of the legs 
should be left without foundation the part supported would still 
maintain its form. More supports are often used than are necessary. 
Thus, if a lathe has two pairs of legs like those shown in B, 
Fig. 243, and these are bolted firmly to the bed, there will be 
four points of support. But if, as suggested by Professor Sweet, 
one of these pairs be connected to the bed by a pin so that the 
support and the bed are free to move relatively to each other 
about the pin, as in Fig. 246, then this is equivalent to a single 
support, and the bed will have three points of support, and will 
maintain its form independently of any change in the foundation. 
This is of special importance when the machines are to be placed 
upon yielding floors. 

220. Reducing Number of Supports. — Fig. 247 shows another 




Fig. 246. Fig. 247. 



case in which the number of supports may be reduced without 
sacrifice. In A three pairs of legs are used. There are therefore 



358 



MACHINE DESIGN. 



six points of support. In B two pairs of legs are used and one 
may be connected by a pin, and there will be but three points of 
support. The chance of the bed being strained from changing 
foundation has been reduced from 6 in ^ to o in 5. The total 
length of bed is 12 feet, and the unsupported length is 6 feet 
in both cases. 

221. Further Correct Methods. — Figs. 248 and 249 show 
correct methods of support for small lathes and planers, due to 





Fig. 248. 



Fig. 249. 



^Professor Sweet. In Fig. 248 the lathe "head -stock " has its 
outlines carried to the foundation by the box pillar; a represents 
a pair of legs connected to the bed by a pin connection, and 
instead of being placed at the end of the bed it is moved in some- 
what, the end of the bed being carried down to the support by a 
parabolic outline. The unsupported length of bed is thereby 
decreased, the stress on the bed is less, and the bed will maintain 
its form regardless of any yielding of the floor or foundation. 
In Fig. 249 the housings, instead of resting on the bed as is usual 
in small planers, are carried to the foundation, forming two of 
the supports; the other is at a and has a pin connection with the 
bed, which being thus supported on three points cannot be twisted 
or flexed by a yielding foundation. 



CHAPTER XX. 



MACHINE FRAMES. 



222. Open-side Frame. — Fig. 250 shows an open -side frame, 
such as is used for punching and shearing machines. During the 
action of the punch or shear a force is apphed to the frame tending 
to separate the jaws. This force may be represented in magnitude, 
direction, and hne of action by P. It is required to find the 
resuhing stresses in the three sections AB^ CD, and EF. Con- 
sider AB. Let the portion above this section be taken as a free 
body. The force P, Fig. 251, and the opposing resistances to 




Fig. 250. 

deformation of the material at the section AB^ are in equihbrium. 
Let H be the projection of the gravity axis of the section AB, 
perpendicular to the paper. Two equal and opposite forces, Pi 
and P2, may be apphed at H without disturbing the equihbrium. 
Let Pi and P2 be each equal to P, and let their line of action be 
parallel to that of P. The free body is now subjected to the 
action of an external couple, PI, and an external force, Pi. The 

359 



360 



MACHINE DESIGN. 



couple produces flexure about H, and the force Pi produces tensile 
stress in the section AB. The flexure results in a tensile stress 
varying from a maximum value in the outer fiber at A to zero 
at H, and a compressive stress varying from a maximum in the 
outer fiber at B to zero at H. This may be shown graphically 
at JK. The ordinates of the line LM represent the varying 
stress due to flexure; while ordinates between LM and NO 
represent the uniform tensile stress. This latter diminishes the 
compressive stress at B, and increases the tensile stress at A. 
The tensile stress per square inch at A therefore equals /+/i; 
where / equals the unit fiber stress due to flexure at ^,and /i 




Fig. 252. 



Pic P 

equals the unit tensile stress due to Pi. Now f =—r , and /i =7-; 

in which c =the distance from the gravity axis to the outer fiber = 
AH, and / = the moment of inertia of the section about H, and 
A =area of the cross-section AB. 

223. Problem. — Let it be required to design the frame of a 
machine to punch f-inch holes in J-inch steel plates, 18 inches 
from the edge. The surface resisting the shearing action of the 
punch = 7rX|''xy =1.18 square inch. The ultimate shearing 
strength of the material is say 50,000 lbs. per square inch. The 
total force P, which must be resisted by the punch frame = 50,000 
X 1.18 = 59,000 lbs. 



MACHINE FRAMES. 361 

The material and form for the frame must first be selected. 
The form is such that forged material is excluded, and difficulties 
of casting and high cost exclude steel casting. The material, 
therefore, must be cast iron. Often the sam.e pattern is used 
both for the frame of a punch and shear. In the latter case, 
when the shear blade begins and ends its cut, the force is not 
apphed in the middle plane of the frame, but considerably to one 
side, and a torsional stress results in the frame. Combined torsion 
and flexure are best resisted by members of box form. The frame 
will therefore be made of cast iron and of box section. The dimen- 
sion AB may be assumed so that its proportion to the ''reach" 
of the punch appears right; the width and thickness of the cross- 
section may also be assumed. From these data the maximum 
stress in the outer fiber may be determined. If this is a safe 
value for the material used the design will be right. 

Let the assumed dimensions be as shown in Fig. 252. Then 

A =bidi — ^2^2 = 78 square inches. 
J _bidi^ — b2d2^ 
12 
= 3002 bi-quadratic inches. 

c=di^2=g^^; /= the reach of the punch +c= 27"; P =59,000 lbs., 
as determined above. Then 

P 59000 
/i=:4-^^ = 756, 

, Pic ^9000X27X9 

' I 3002 ^1 ' f 

h +/ "^5532 =maximum stress in the section. 

The average strength of cast iron such as is used for machinery 
castings, is about 20,000 lbs. per square inch. The factor of 
safety in the case assumed equals 20,000-^5532=3.65. This is 
too small. There are two reasons why a large factor of safety 



362 



MACHINE DESIGN. 



should be used in this design: I. When the punch goes through 
the plate the yielding is sudden and a severe stress results. This 
stress has to be sustained by the frame, which for other reasons 
is made of unresilient material. II. Since the frame is of cast 
iron, there will necessarily be shrinkage stresses which the frame 
must sustain in addition to the stress due to external forcels. 
These shrinkage stresses cannot be calculated and therefore can 
only be provided against by a large factor of safety. 

Cast iron is strong to resist compression and weak to resist 
tension, and the maximum fiber stress is tension on the inner 
side. The metal can therefore be more satisfactorily distributed 
than in the assumed section, by being thickened where it sustains 
tension, as at ^, Fig. 253. If, however, there is a very thick 
body of metal at a, sponginess and excessive shrinkage would 
result. The form B would be better, the metal being arranged 
for proper cooling and for the resisting of flexure stress. 






Fig. 253. 



Fig. 254. 



Dimensions may be assigned to a section like B and the 
cross-section may be checked for strength as before. See Fig. 254. 
GG, a Hne through the center of gravity of the section, is found to 
be at a distance of 7.05 inches from the tension side.* The 



* A simple and satisfactory method for obtaining a close approximation to 



MACHINE FRAMES. 363 

required values are as follows: (7 = 7.05 inches; / = reach of 
punch +c = i8 + 7. 05 =25.05 inches; ^=156.25 square inches; 
/ = 5032.5 bi-quadratic inches ; P = 59,000 lbs. 

Then /,=^^59ooo^ 

A 156.25 ^'' ' 

Pic 59000X25.05X7.05 

7 = ^^ = = 2070.4 lbs. 

' I 5032-5 ' 

/i4-/ = 2448 lbs. =maximum fiber stress in the section. The 
factor of safety =20,000 -^- 2448 =8.17.* This section, therefore, 
fulfills the requirement for strength, and the material is well 
arranged for cooling with little shrinkage, and without spongy 
spots. The gravity axis may be located, and the value of / deter- 
mined by graphic methods. See Hoskins's ''Graphic Statics." f 
Let the section CD, Fig. 250, be considered. Fig. 255 shows 
the part at the left of CD free. K is the projection of the gravity 
axis of the section. As before, put in two opposite forces, Pg 
and P4, equal to each other and to P, and having their common 
line of action parallel to that of P, at a distance h from it. P 
and P4now form a couple, whose moment =P/i, tending to produce 
flexure about K. P3 must be resolved into two components, 
one P37, at right angles to the section considered, tending to 
produce tensile stress; and the other JK, parallel to the section, 
tending to produce shearing stress. The greatest unit tensile 

the true gravity axis of irregular figures is as follows: On a -piece of thin but uni- 
form cardboard lay out the figure to scale. Cut it out carefully with a sharp 
knife. Balance the figure exactly, by trial, on a knife-edge. The line of contact 
with the knife-edge is the gravity axis. Its position may be marked and its loca- 
tion measured to scale. 

* This discussion neglects the action of gravity which would exert a counter- 
balancing moment, reducing the maximum tensile fiber stress below the value 
found. This makes the actual factor of safety greater than the apparent {actor 
of safety. 

t The student will be familiar with analytical methods for their determina 
tion from his studv of the 'Mechanics of Materials." 



364 



MACHINE DESIGN. 



stress in this section will equal the sum of that due to flexure 
and that due to tension 



The greatest unit shear =/« = 



JK 

A ' 



In the section FE, Fig. 250, which is parallel to the line of 
action of P, equal and opposite forces, each=P, may be intro- 
duced, as P5 and Pq. P and Pq will then form a couple with an 




Fig. 256. 



arm /2, and P5 will be wholly appHed to produce shearing stress. 
The maximum unit tensile stress in this section will be that due 
to flexure, j = Pl2C-^I, and the maximum unit shear will be 
js = P^A. Any section may be thus checked. 



MACHINE FRAMES. 



365 



The dimensions of several sections being found, the outline 
curve bounding them should be drawn carefully, to give good 
appearance. The necessar}^ modifications of the frame to pro- 
vide for support, and for the constrainment of the actuating 
mechanism, may be worked out as in Fig. 256. A is the pinion 
on the pulley shaft from which the power is received; B is the 
gear on the main shaft ; C, D, and G are parts of the frame added 
to supply bearings for the shafts ; E furnishes the guiding surfaces 
for the punch "slide." The method of supporting the frame is 
shov^Ti, the support being cut under at F for convenience to the 
workman. The parts C, Z), £, and G can only be located after 
the mechanism train has been designed. 




Fig. 257. 

224. Slotting-machine Frame. — See Fig. 257. It is specified 
that the slotter shall cut at a certain distance from the edge of 
any piece, and the dimension ^iJ is thus determined. The 
table G must be held at a convenient height above the floor, and 
RK must provide for the required range of "feed." K is cut 
under for convenience of the workman, and carried to the floor 
line as shown. It is required to "slot " a piece of given vertical 
dimension, and the distance from the surface of the table to E 
is thus determined. Let the dimension LM be assumed so that 



366 MACHINE DESIGN. 

it shall be in proper proportion to the necessary length and height 
of the machine. The curves LS and MT may be drawn for 
bounding lines of a box frame to support the mechanism. M 
should be carried to the floor line as shown, and not cut under. 
None of the part DNE, nor that which serves to support the 
cone and gears on the other side of the frame, should be made 
flush with the surface LSTM, because nothing should interfere 
with the continuity of the curves LS and TM. The supporting 
frame of a machine should he clearly outlined, and other parts should 
appear as attachments. The member VW should be designed so 
that its inner outline is nearly parallel to the outline of the cone 
pulley, and should be joined to the main frame by a curve. The 
outer outline should be such that the width of the member increases 
slightly from W to F, and should also be joined to the mai j 
frame by a curved outline. In any cross-section of the frame, 
as XX, the amount of metal and its arrangement may be con- 
trolled by the core. It is dictated by the maximum force, P, 
which the tool can be required to sustain. The tool is carried by 
the slider of a slider-crank chain. Its velocity varies, therefore, 
from a maximum near mid-stroke, to zero at the upper and lower 
ends of its stroke. The belt which actuates the mechanism runs 
on one side of the steps of the cone pulley, at a constant velocity. 
Suppose that the tool is set (accidentally) so that it strikes the 
table just before the slider has reached the low^er end of its stroke. 
The resistance, R, offered by the tool to being stopped, multiplied 
by its (very small) velocity, equals the difference of belt tension 
multiplied by the belt velocity (friction and inertia neglected).* 
R, therefore, would vary inversely as the slider velocity, and hence 
may be very great. Its maximum value is indeterminate. A 
"breaking piece " may be put in between the tool and the crank. 
Then when R reaches a certain value, the breaking piece fails. 

* See Chapter V. 



MACHINE FRAMES. 367 

The stress in the stress-members of the machine is thereby limited 
to a certain definite value. From this value the frame may be 
designed. Let P = upward force against the tool when the break- 
ing piece fails.* Let / = the horizontal distance from the line of 
action P to the gravity axis of the section XX. Then the section 
XX sustains flexure stress caused by the moment PI, and tensile 
stress equal to P, The maximum unit stress in the section 

^ ^ Pic P 

A section may be assumed and checked for safety, as for the 
punching-machine in § 223. 

225. Stresses in the Frame of a Side-crank Steam-engine. — 

Fig. 258 is a sketch in plan of a side -crank engine of the ''girder 




bed '^ type. The supports are under the cylinder C, the main 
bearing £, and the out-board bearing D. A force P is applied 
in the center line of the cylinder, and acts alternately toward the 
right and toward the left. In the first case it tends to separate 
the cylinder and main shaft; and in the second case it tends to 
bring them nearer together. The frame resists these tendencies 
with resulting internal stresses. 

Let the stresses in the section AB be considered. The end of 
ths frame is shown enlarged in Fig. 259. If the pressure from the 
piston is toward the right, the stresses in AB will be: I. Flexure 

* P is limited to the friction due to screwing up the four bolts which hold the 
tool. 



368 



MACHINE DESIGN. 



due to the moment PI, resulting in tensile stress below the gravi'ty 
axis, N, with a maximum value at h, and a compressive stress 
above N with a maximum value at a. II. A direct tensile stress, 
= P, distributed over the entire section, resulting in a unit stress = 
P-^A =ji lbs. per square inch. This is shown graphically at n, 
Fig. 259. aibi is a datum line whose length equals ab. Tensions 
are laid off toward the right and compressions toward the left. 

Jcieiai 



^ 



L 



6a cidi 
(n) 



a 


ieoJci A 


X 




^Ns N- 


// 


i 1 


d2C2bo ' 

(m) B 


p 



Fig. 259. 



The stress due to flexure varies directly as the distance from the 
neutral axis Ni, being zero at Ni. If, therefore, biCi represents 
the tensile stress in the outer fiber, then ciki drawn through Ni 
will be the locus of the ends of horizontal lines, drawn through 
all points of aibi, representing the intensity of stress in all parts 
of the section, due to flexure. If cidi represent the unit stress 
due to direct tension, then, since this is the same in all parts of 
the section, it will be represented by the horizontal distance be- 
tween the parallel lines ciki and diei. This uniform tension 
increases the tension biCi due to flexure, causing it to become b^di ; 
and reduces the compression ^i^i, causing it to become eiai. 
The maximum stress in the section is therefore tensile stress in 
the lower outer fiber, and is equal to bidi. 

When the force P is reversed, acting toward the left, the 
stresses in the section are as shown at m, Fig. 259: compression 
due to flexure in the lower outer fiber equal to C2&2; tension due 
to flexure in the upper outer fiber equal to ^2^2; and uniform 
compression over the entire surface equal to ^2^2- This latter 
increases the compression in the lower outer fiber from ^2<^2 to 
&2^2> and decreases the tension in the upper outer fiber from ^2^2 



MACHINE FRAMES. 



369 



to 02^2- The maximum stress in the section is therefore compres- 
sion in the lower outer fiber equal to 62^2- The maximum stress, 
therefore, is always in the side of the frame next to the connecting- 
rod. 

If the gravity axis of the cross-section be moved toward the 
connecting-rod, the stress in the upper outer fiber will be increased, 
and that in the lower outer fiber will be proportionately decreased. 
The gravity axis may be moved toward the connecting-rod by 
increasing the amount of material in the lower part of the cross - 
section and decreasing it in the upper part. 

The stress in any other section nearer the cylinder will be due 
to the same force, P, as before; but the moment tending to pro- 
duce flexure will be less, because the lever arm of the moment is 
less and the force constant. 

226. Heavy-duty Engine Frame. — Suppose the engine frame 
to be of the type which is continuous with the supporting part as 
shown in Fig. 260. Let Fig. 261 be a cross-section, say at AB. 
O is the center of the cylinder. The force P is appHed at this 



Fig. 260. 



E^ 




Fig. 261. 



point perpendicular to the paper. C is the center of gravity of 
the section (the intersection of two gravity axes perpendicular 
to each other, found graphically). Join C and O, and through 



Pic 


in known tenns. 


p 


in known tenns. 


^ area of section' 



370 MACHINE DESIGN. 

C draw XZ perpendicular to CO. Then XX is the gravity 
axis about which flexure will occur.* The dangerous stress will 
be at F, and the value of c will be the perpendicular distance from 
F to XX. The moment of inertia of the cross-section about 
.YX may be found, =7; /, the lever-arm of P, =0C. The 
stress at F, j + ji must be of safe value. 



227. Closed Frames. — Fig. 262 shows a closed frame. The 
members G and H are bolted rigidly to a cylinder C at the top, and 
to a bedplate, DD, at the bottom. A force P may act in the center 
line, either to separate D and C, or to bring them nearer together. 
The problem is to design G, H, and D for strength. If the three 
members were ''pin connected" (see Fig. 263), the reactions of 
C upon A and B at the pins would act in the hues EF and GH. 
Then if P acts to bring D and C nearer together, compression 
results in A, the line of action being EF; compression results 
in B, the line of action being GH. These compressions being in 
equilibrium with the force P, their magnitude may be found by 
the triangle of forces. From these values A and B may be designed, 
C is equivalent to a beam whose length is /, supported at both 

* This is not strictly true. If OC is a diameter of the "ellipse of inertia," 
flexure will occur about its conjugate diameter. If the section of the engine frame 
is symmetrical with respect to a vertical axis, OC is vertical, and its conjugate 
diameter XX is horizontal. Flexure would occur about XX, and the angle be- 
tween OC and XX would equal 90°. As the section departs from symmetry 
about a vertical, XX, at right angles to OC, departs from OC's conjugate, and 
hence does not represent the axis about which flexure occurs. In sections like 
Fig. 259, the error from making ^=90° is unimportant. When the departure 
from sjonmetry is very great, however, OC's conjugate should be located and 
used as the axis about which flexure occurs. For method of drawing " ellipse of 
inertia" see Hoskins's '* Graphic Statics." 



MACHINE FRAMES. 



371 



ends, sustaining a transverse load P, and tension equal to the 
horizontal component of the compression in A or B. The data 
for its design would therefore be available. Reversing the direc- 
tion of P reverses the stresses ; the compression in A and B becomes 
tension; the flexure moment tends to bend C convex downward 
instead' of upward, and the tension in C becomes compression. 
But when the members are bolted rigidly together, as in 

Fig. 262. 




Fig. 263. 



Fig. 262, the lines of the reactions are indeterminate. Assump- 
tions must therefore be made. Suppose that G is attached to D 
by bolts at E and A. Suppose the bolts to have worked sHghtly 
loose, and that P tends to bring C and D nearer together. There 
would be a tendency, if the frame yields at all, to relieve pressure 
at E and to concentrate it at ^. The Hne of the reaction would 
pass through A and might be assumed to be perpendicular to 



372 MACHINE DESIGN. 

the surface AE. Suppose that P is reversed and that the bolts 
at A are loosened, while those at E are tight. The line of the 
reaction would pass through E, and might be assumed to be 
perpendicular to EA. MN is therefore the assumed line of the 
reaction, and the intensity i?=P-^2. In any section of G, as 
XX, let K' be the projection of the gravity axis. Introduce at 
K' two equal and opposite forces equal to R and with their lines 
of action parallel to that of R. Then in the section there is flex- 
ure stress due to the flexure moment Rl, and tensile stress due to 
the component of R2 perpendicular to the section, =Rz. Then 
the maximum stress in the section =/+/i. 

,_i?3 , _RIC 

f-A'^ f'~ I' 

A section may be assumed, and A, I, and c become known; 
the maximum stress also becomes known, and may be compared 
with the ultimate strength of the material used. 

Obviously this resulting maximum stress is greater when the 
line of the reaction is MN than when it is KL. Also it is greater 
when MN is perpendicular to EA than if it were inclined more 
toward the center line of the frame. The assumptions therefore 
give safety. If the force P could only act downward, as in a 
steam hammer, KL w^ould be used as the line of the reaction. 

The part D in the bolted frame is not equivalent to a beam 
with end supports and a central load like C, Fig. 263, but more 
nearly a beam built in at the ends with central load, and it may 
be so considered, letting the length of the beam equal the hori- 
zontal distance from E to F,=li. Then the stress in the mid- 

Ph 
section will be due to the flexifre moment -^, and the maximum 

o 

PI C 

stress =/ = -^^. The values c and / may be found for an assumed 
section, and / becomes known. 



MACHINE FRAMES. 373 

228. Steam-hammer Frames. — Steam-hammers are made both 
with "open-side " and "closed " frames. They may therefore be 
designed by methods already given, if the maximum force applied 
is known. The problem is, therefore, to find the value of this 
maximum force. 

There are two types of steam-hammers: 

Type I. Single-acting. A heavy hammer-head attached to 
a steam-piston is raised to a certain height by steam admitted 
under the piston. The steam is then exhausted and the hammer- 
head with attached parts falls by gravity to its original position. 
The energy of the blow = Wl, where W is the falling weight and 
/ is the height of fall. 

Type 2. Double-acting. A lighter hammer-head is lifted by 
steam acting under its attached piston, and during its fall steam 
is admitted above the piston to help gravity to force it downward. 
The energy of the blow •= Wl (as before) plus the energy received 
from the expansion of the steam; or, if the steam acts throughout 
the entire stroke, the energy of blow = Wl + pAl, where p is the 
mean pressure per square inch and A is the area of the upper 
side of the piston. 

229. Stresses in Single-acting Frames. — In type i, when the 
action is as described, a force acts downward upon the frame during 
the lifting of the hammer. The intensity of this force =pA =the 
mean pressure of steam admitted multiplied by area of piston, 
and the line of action is the axis of the piston-rod. During the 
fall of the hammer the cylinder and frame act simply as a guide, 
and no force is applied to the frame except such as may result 
from frictional resistance. The hammer strikes an anvil which 
is not attached to the frame, but rests upon a separate foundation. 

But a greater force than pA may be applied to the frame. 
In order that a cushioned blow may be struck, the design is such 
that steam may be introduced under the piston at any time during 
its downward movement, and this steam is compressed by the 



374 



MACHINE DESIGN. 



advancing piston. A part of the energy of the falling hammer 
is used for this compression. The pressure in the cylinder result- 
ing from this compression is communicated to the lower cylinder- 
head and through it to the frame. Under certain conditions steam 
might be admitted at such a point of the stroke that all of the 
energy of the falHng hammer might be used in compressing the 
steam to the end of the stroke. The hammer would then just 
reach the anvil, but would not strike a blow. 

Fig. 264, a shows by diagram a hammer of type i. Steam is 
admitted, the piston is raised, the exhaust- valve is opened, and 




'bt y 


1 


n 


R 


(6-) 

L K 




N 


Q "'^■^--— -_ ^M 


F- 


1 



/ \ 



Fig. 264. 



the piston falls. But at some point in the stroke steam is again 
admitted, filling the cylinder, and the valve is closed. Com- 
pression occurs and absorbs all or part of the energy WL In the 
latter case the hammer will strike the anvil a blow whose energy 
is equal to Wl minus the work of compressing the steam in C 



MACHINE FRAMES. 375 

The compression is shown upon a pressure -volume diagram, 

Fig. 264, h. Progress along the vertical axis from B toward E 

corresponds to the downward movement of the hammer. Vertical 

ordinates therefore represent space, 5, moved through by the 

hammer; or, since SA = volume displaced by the piston, the 

vertical ordinates may also represent volumes. EF represents 

V2 
the volume V2 of the clearance space, or -r, the piston movement 

which corresponds to the clearance. Horizontal ordinates meas- 
ured from BF represent absolute pressures per square inch. 
Let pi represent the absolute boiler pressure represented by DK. 
TN is the line of atmospheric pressure. During the lifting of the 
hammer the upper surface of the piston is exposed to atmospheric 
pressure and the lower surface is exposed to pressure just suffi- 

W 
cient to raise the hammer, =-j-. The work of Hf ting is represented 

by the area NTHG. This work equals the energy, Wl, which 
the hammer must give out in some way before reaching the anvil 
again. When the piston has fallen to some point, as Z>, steam 
may be let in below it at boiler pressure, DK. The advancing 
piston will compress this steam, and KM will be the compression 
curve.* The work of compression is represented by the area 
RKMN. If the compression is to absorb all the energy Wl, the 
area which represents the work of compression must equal the 
area which represents Wl. Hence area NTHG must equal 
RKMN: or, since the area RLGN is common to both, the area 
RTHL must equal the area LKMG. The point at which com- 
pression must begin in order to cause this equality may be found 
by trial. The greatest unit pressure reached by compression is 
represented by EM. The greatest pressure, ^2, upon the lower 
cylinder-head is represented by NM^ since atmospheric pressure 



* Assuming pV = constant. 



2>1^ MACHINE DESIGN. 

acts on the outside. The corresponding total force communicated 
to the frame =p2A =P * 

If compression had begun earher the energy would have been 
absorbed before the hammer reached the anvil, the piston would 
have stopped short of the end of the stroke, the compression curve 
would have been incomplete, and the greatest pressure would have 
been less than EM. Obviously if compression had begun later 
the greatest pressure would have been less than EM. Therefore 
the force P, =p2A, with the cylinder's' axis for its line of action, 
is the greatest force that can be appHed to the frame in the regular 
working of the ham^mer. 

A greater for.ce might be accidentally applied. For, suppose 
that water is introduced into the cylinder in such quantity that 
the piston reaches it before the hammer reaches the anvil, then 
all the energy will be given out to overcome the resistance of the 
water. The resulting force is indeterminate, because the space 
through which the resistance acts is unknown. This force may 
be very great. The force applied to the frame may be limited 
by the use of a "breaking-piece." Thus the studs which hold 
on the lower cylinder-head may be drilled f so that they will 
break under a force KP, in which i^ is a factor ot safety and P 
is the force found above. Then the breaking-piece will be safe 
under the maximum working force, but will yield when an acci- 
dental force equals KP, thus limiting its value. The frame may 
be designed for a maximum force KP. 

230. Problem, Type i. — Let W, weight of hammer and 
attached parts, =2000 lbs.; /, maximum length of stroke, =24 
inches; A, effective area of piston, =50 square inches; clearance 
= 15 per cent; boiler pressure =85 lbs. by gauge. Steam is 
admitted to Hit the hammer, pressure being controlled by throtthng. 

* In which A is the efifective area of the piston, i.e., area of the piston less 
area ot the rod. 
t See page 136. 



MACHINE FRAMES. 377 

The pressure per square inch that will just lift the hammer = 
2000 lbs. -^ 50 square inches =40 lbs. In Fig. 264, NG repre- 
sents 40 lbs., and NT represents the volume displaced by the 
piston during a complete stroke. Hence NTHG represents the 
work of Hfting the hammer, or the energy that must be absorbed 
just as the hammer reaches the anvil. Trial shows that to 
accomplish this, compression must begin at just about 6 inches 
from the end of the stroke. The maximum resulting pressure, 
represented by ATikf, equals 258 lbs. per square inch. The total 
pressure acting downward on the frame =^2^ = 258 X 50 = 12,900 
lbs. =P. If the factor of safety, K, is 5, the strength of the 
breaking-piece =XP = 5X12,900 =64,500 lbs. This is the maxi- 
mum force, and hence may be used as a basis of the frame design. 

231. Stresses in Double-acting Frames. — In type 2 the 
maximum working force may be found by a similar method. In 
Fig. 265, NG represents the pressure per square inch of piston 
necessary to raise the hammer. The area NTHG represents 
the energy stored in the hammer by lifting. The area HSJL 
represents the work done by steam at boiler pressure acting on 
the upper piston face while the piston descends to D. At this 
point steam is exhausted above the piston and let in below it, and 
compression takes place during the remainder of the stroke. To 
absorb ail the energy of the hammer by compression, the areas 
NTSJLG and RKMN must be equal. The area NRLG is 
common to both; hence the area LKMG must equal the area 
RTSJ. The point at which compression must begin in order 
to cause this equality may be found by trial. 

232. Problem, Type 2. — Let PT, weight of hammer and 
attached parts, =600 lbs.; /, maximum length of stroke, =24 
inches; A, effective area of piston (both faces), =50 square 
inches; clearance = 15 per cent; boiler pressure = 85 lbs. by gauge. 
The construction in Fig. 265 shows that compression, beginning 
at 9J inches before the end of the piston's stroke, absorbs all 



378 



MACHINE DESIGN. 



the energy of the hammer, and gives 325 lbs. as a maximum 
pressure per square inch. Then the maximum working force 
= 325 X 50 = 16,250. If i^ = 5, the strength of the breaking-piece 
= 16,250X5 =81,250 lbs. 




m 



D 



Fig. 265. 

233. Other Stresses in Hammer Frames. — ^An accidental force 
acting upward may be applied to the hammer frame. The boiler 
pressure is necessarily greater than that which is necessary to 
lift the hammer.* Thus in § 230 a pressure of 40 lbs. per square 
inch is sufficient to lift the hammer, but the boiler pressure is 
85 lbs. per square inch. If the throttle-valve were opened wide 
and held open during the movement of the hammer upward, the 
energy stored in the hammer when it reaches its upper position 
would equal the product of boiler pressure, piston area, and length 
of stroke, =85X50X24 = 106,000 in.-lbs. The energy necessary 



* So that it may be possible to work the hammer when the steam-pressure is 
lower in the boiler. 



M/i CHINE FRAMES. 379 

to just lift the hammer is 40X50X24=48,000 in. -lbs. The 
difference between these two amounts of energy, =58,000 in. -lbs., 
will exist as kinetic energy of the moving hammer; and it must 
be absorbed before the hammer can be brought to rest in its 
upper position. The force which would result from stopping 
the hammer would be dependent upon the space through which 
the motion of the hammer is resisted. Springs are often provided 
to resist the motion of the hammer when near its upper position. 
These springs increase the space factor of the energy to be given 
out and thereby reduce the resulting force. An automatic 
device for closing the throttle -valve before the end of the stroke 
and introducing steam for compression above the piston may 
be used. The steam is then a fluid spring. 

234. Design of Crane Frames. — A crane frame is to be de- 
signed from the following specifications : Maximum load, 5 tons 
= 10,000 lbs.; radius = maximum distance from the line of lifting 
to the axis of the mast, =18 feet; height of mast = 20 feet; radial 
travel of hook in its highest position = 5 feet; axis of jib to be 15 
feet above floor line. Fig. 266 shows the crane indicated by the 
center lines of its members. 

The external forces acting on the crane may be considered first. 
A load of 10,000 lbs. acts downward in the line ah. This is held 
in equihbrium by three reactions : one acting horizontally toward 
the left through the upper support, i.e., along the line hc\ another 
acting horizontally toward the right through the lower support, 
i.e., in the hne ad\ a third acting vertically upward at the lower 
support, i.e., in the line cd. The crane is a ''four-force piece." 
One force, AB, is completely known, the other three are known 
only in line of action. Produce ah and he to their intersection at 
M. The line of action of the resultant of ah and he must pass 
through M. The resultant of cd and da must be equal and 
opposite to the resultant of ah and he, and must have the same 
line of action. But the line of action of the resultant of cd and 



38o 



MACHINE DESIGN. 



da must pass through N. Hence MN is the common line of 
action of the resuhants of ah and he and of cd and da. Draw 
the vertical line AB * representing 10,000 lbs. upon some assumed 
scale ; from B draw BC parallel to he, and from A draw ylC parallel 
to MN. The intersection of these two hnes locates C and deter- 
mines the magnitude of BC. Now ^C is the resultant oi AB 
and BC, and CA, equal and opposite, is the resultant of CD 




Force Diagram 



• Fig. 266. 

and DA. Therefore CA has but to be resolved into vertical and 
horizontal components to determine the magnitudes of CD and 
DA. The force polygon is therefore a rectangle and CD=ABy 
ziidBC^DA. 

From the forces AD and CD acting at -N, the supporting 
journal and bearing at the base of the crane may be designed; 
and from the force BC, acting at V, the upper journal and bearing 
may be designed. 

235. Jib. — The forces acting on the jib are, first, AB acting 
vertically downward at its end; second, an upward reaction 

* See Force Diagram, Fig. 266. 



MACHINE FRAMES. 



381 



from the brace at iJ, whose line of action coincides with the axis 
of the brace;* third, a downward reaction at L where the jib 
joins the mast, whose Hne of action must coincide with the Hne 
of action of. the resultant of AB and the brace reaction. LK is 
therefore this line of action. 

Fig, 267. 




.<'?^ 



Fig. 268. 

In the force diagram draw BE parallel to the center line of the 
brace, and draw EA parallel to LK. Then BE will represent the 
brace reaction, and EA will represent the reaction at L in the Hne 
ae. Let RQ, Fig. 267, represent the jib isolated, ae^ he, and ah 
are the lines of action of the three forces acting upon it. The 
vertical components of these forces are in equilibrium, and tend 
to produce flexure in the jib. The horizontal components are 
in equilibrium and tend to produce tension in the jib. The vertical 



* Considering the joint between the brace and jib equivalent to a pin con- 
nection. 



382 MACHINE DESIGN. 

force acting at R is jP^,* =5000 lbs., the vertical force acting at 
T is FB, =15,000 lbs., and the vertical force acting at Q is ABj 
= 10,000 lbs. 

Flexure is also produced in the jib by its own weight acting as 
a uniformly distributed load. 

In order to design the jib, standard rolled fomas may be 
selected which will afford convenient support for the sheave car- 
riage. Two channels located as shown in Fig. 268 will serve for 
this crane. For trial 12 -inch heavy channels are chosen. From 
Carnegie's Hand-book, the moment of inertia for each channel 
about an axis perpendicular to the web at the center =7 = 248; 
f = 6 inches; the weight, w, of two channels per inch of length 
= 81 lbs. 

The total weight of the two channels =w/ = 8JX 18 X 12 = 
1800 lbs. The vertical, reaction, Pi, at R, Fig. 267, due to this 

weight is Pi = ( —■ I -^ I2, from the equation of moments, 

due to the weight of jib about the point T. Introducing numerical 
values. Pi =450 lbs. The total reaction at R is therefore 5000 
— 450=4550 lbs. A diagram of moments of flexure may now be 
drawn under the jib. Fig. 267. Considering the portion PQ, the 

moment at P = P/i+ =741,600 in. -lbs. Divide TQ into 

three equal parts. At the division nearest T the moment = 

Ph+ ; in which ^3= the distance from Q to the section con- 

2 

sidered. The moment at the other two points may be found 

by similar method. 

The moment at any pomt at the leit of r = 455oX/4H ; 

in w^hich U is the distance from R to the section considered. From 
the values thus found the diagram of flexure moments may be 

* See force diagram, Fig. 266. 



MACHINE FRAMES. 



383 



drawn. The maximum value is at T, where M = 741,600 in. -lbs. 



The resultinoj fiber stress 



Mc 741600X6 



^ = -j = 



= 8971 lbs. The 



248X2 

horizontal component acting at R is equal to FE (see force dia- 
gram, Fig. 266) =12,000 lbs. An equal and opposite horizontal 
force must act at T. Between T and Q there is no tensile stress 
due to the forces AB, BE, and AE. 



Fig. 269. 




Fig. 270. 



Another force which modifies this result needs to be considered. 
Let AB, Fig. 269, be the upper surface of the jib. The load is 
supported as shown. The chain which is fastened at B passes 
over the right-hand carriage sheave, do^vn and under the hook 
sheave, up and over the left-hand carriage sheave, horizontally 
to the sheave at A, and thence to the winding drum. If a load 
of 10,000 lbs. is supported by the hook there will be, neglecting 
friction, a tension of 5000 lbs. throughout the entire length of the 



3^4 MACHINE DESIGN, 

chain from B to the winding drum. There is therefore a force 
of 5000 lbs. tending to bring A and B nearer together, and hence 
to produce compression in the jib * The resultant tension be- 
tween R and T is 12,000-5000 = 7000 lbs., while between T and 
Q there is a compression of 5000 lbs. The cross-sectional area 
of the two channels selected = 30 square inches. Hence the unit 
tensile stress =/i =7000^-30 = 233 lbs. The maximum unit ten- 
sile stress in the jib =/ + /i =8971 4-233 =9204 lbs. per square inch. 

If the channels are of steel, their unit tensile strength will 
probably equal 60,000 lbs. per square inch. The factor of 
safety =60,000-^9204 = 6.5. In a crane a load may drop through 
a certain space by reason of the slipping of a link that has been 
caught up, or the failure of the support under the load while the 
chain is slack. When this occurs a blow is sustained by the stress 
members of the crane. The energy of this blow equals the load 
multiplied by the height of fall. But the stress members of the 
crane are long, and the yielding is large. Hence the space through 
which the blow is resisted is large and the resulting force is less 
than with small yielding. In other words, the stress members act 
as a spring, reducing the force due to shock. Hence, in a crane 
of this type, the ductile and resilient material is liable to modified 
shock, and a factor of safety = 6.5 is large enough. 

The jib might also be checked for shear, but in general it will 
be found to have large excess of strength. 

236. Mast. — Fig. 270 shows the mast by its center line with 
the Hues of action of the forces acting upon it. It is equivalent 
to a beam supported at C and D with a load at A. The moment 
of flexure at A equals the force acting in the line hc^ multiplied 
by the distance CA in inches, =9000 lbs. X 60 inches = 540,000 



* There is also flexure due to this force multiplied by the distance from the 
centre line of the horizontal chain to the gravity axis of the jib. This is small 
and may be neglected. 

t The force BC in Fig. 266. 



MACHINE FRAMES. 385 

in. -lbs. =M. The flexure moment is a maximum at this point. 

and decreases uniformly toward both ends. The moment diagram 

is therefore as shown. The maximum fiber stress due to this 

flexure moment = / = Mc ^ I. Selecting two light 1 2 -inch channels, 

T A A • u ; 540000X6 

/ = 1 76 ; c=6 mches ; / = — — — t~ = 9200 lbs. 

The tension in the mast equals the vertical component of the 
force acting in the line 6^^,* = 5000 lbs. (Actually reduced to 
4550 by the effect of the weight of the jib.) The compression in 
the mast due to the tension in the chain = 5000 lbs. between A 
and the point of support of the winding drum B. The tension 
and compression therefore neutrahze each other, except below 5, 
where the flexure moment is small. Hence the maximum unit 
stress in the mast is 9200 lbs. The factor of safety =60000-^9200 
= 6.5, which is safe as before. This also may be checked for shear. 

237. Brace. — The compression stress in the brace is 19,000 
lbs.,t and the length, 19 feet, is such that is needs to be treated 
as a "long column." Because of the yielding of joints and of 
the other stress members, the brace is intermediate between a 
member with "hinged ends" and "flat ends"; therefore for safety 
it should be considered as hinged. In the treatment of long 
columns, the " straight -line formula" will be used.t This 
formula is of the form 

- = p = B-C-. 
A ^ r 

P is the total force that will cause incipient buckling, and hence 
the force that will destroy the column; A is the cross-sectional 
area of the column; p is the unit stress that will cause buckling; 

* The force AE in Fig. 266. 

•J- See force diagram, Fig. 266. 

X For discussion of long-column formulae see "Theory and Practice of Mod- 
ern Framed Structures," by Johnson, Bryan, and Turneaure, page 143. Published 
by John Wiley & Sons. 



3^6 MACHINE DESIGN. 

B and C are constants derived from experiments on long columns 
(the values of B and C vary with the method of attachment of the 
ends of the column, and with the material of the column) ; / is 
the length of the column in inches; and r is the radius of gyration 
of the cross-section, =\/l^A, I being the moment of inertia of 
the cross-section referred to the axis about which buckling takes 
place. 

Values of B and C are as follows: 

P I 

For wrought iron, hinged ends, -j =42000-157- 

P 7 

flat ^ = 42000 — 128- 

p I 

** mild steel, hinged " y = 52500 — 220-. 



*' '' '' flat 



P I 

2 = 52500-179-. 



The brace will be of mild steel channels, and the ends will be 
considered as hinged. The formula to be used is therefore 



from which 



P I 

"1=52500-220-, 



P=(^52500-220-j^. 



Channel bars may be selected and values of r and A become 
known from tables. For trial 5 -inch hght channels are chosen. 
Carnegie's tables give for 2 channels, ^ = 3.8 inches, and A = 
3.9 square inches. Introducing these values in the above equation, 
with / = i9'Xi2 =228'', gives ^ = 153, 270 lbs. Since the maxi- 
mum compression force sustained by the brace =19,000 lbs., 



MACHINE FRAMES. 



387 



the factor of safety = 153,270^-19000=8 + . This is a larger 
value than those for the jib and mast, but it is probably inadvisable 
to use smaller channels because of convenience in making the 
connection with the jib and mast. 

But the brace must be made safe against side buckling. The 
two channels may be considered as acting as a single member if 
they are braced laterally. The lateral bracing will be determined 
later. In Fig. 271 the moment of inertia about the axis X for 



^— '-— i 





, 


Y 


IQ 








■=^ 




< — 


b-* 


1 






<-a^ 










.=3 




<-£G- 


> 








Q 




Fig. 271. 



Fig. 272 



each channel = 7 (from table.) If the moment of inertia of each 
about the axis Y be made = 7, the radius of gyration will be the 
same about both axes, the values in the above equation will be 
the same, and there will be the same safety against side buckling 
as against buckling in the plane through the axis of the mast. 
Therefore it is only necessary to make the distance, a, of each 
channel from the axis of Y such that /y = 7. The moment of 
inertia of one channel about its own gravity axis GG =0.466. Its 
moment of inertia about Y^'j^Iq+Ax^. Solving, x^ = (ly — /<-) 
■^Af whence 



X 



2 = 



7—0.466 



1-95 
a; = 1.828. 



3-35. 



Hence the distance apart of the gravity axes = 1.828X2 =3.65. 
But the gravity axis is 0.44 inch from the face of the web, i.e.^ 
x — a = o.44. Therefore the distance, b, between the channels 
=3.65 — 0.88 = 2.77 inches. Convenience in construction would 



388 MACHINE DESIGN. 

undoubtedly dictate a greater distance, and hence greater safety 
against side buckling. 

The position of these two channels relative to each other must 
be insured by some such means as diagonal bracing. See Fig. 272. 
The distance, /, must be such that the channels shall not buckle 
separately under half the total load. Solving the long-column 
formula gives 



/ = (525oo-f)^^, 



in which P is thie load sustained by each channel ( = 19000-^2 = 
9500 lbs.) multiplied by the factor of safety, say 6. The radius of 
gyration, r, is about a gravity axis parallel to the web, = V/ -^A =* 
V0.466 -M .95 = 0.488. 

, / 9500X6X0.488 

/= 52500-^^ -^^— = 51.8. 

V^ ^ 1.95 / 220 ^ 

The value of /, therefore, must not be greater than 51.8 inches 
but it may be less if convenience, or the use of standard braces 
requires. 

238. Crane Frame with Tension Rods. — The brace in the 
crane just considered may be replaced by tension rods, as shown 
in Fig. 273. This allows the load to be moved radially through- 
out the entire length of the jib. The force polygon. Fig. 274, 
shows tension equal to 37,000 lbs. in the tension rods, and com- 
pression in the jib = 35,800 lbs. If the tension rods are made of 
mild steel with an ultimate tensile strength of 60,000 lbs., and 
a factor of safety =6, the cross -sectional area must equal 
37,000X6-^60,000 = 3.7 square inches. If two rods are used the 
minimum diameter of each = 1.535 ; say, i^ inches. 

The mast is a flexure member 20 feet long supported at the 
ends, and sustaining a transverse force of 35,800 lbs. at a distance 
of 5 feet from the upper end. The upper end reaction is there- 



MACHINE FRAMES. 



389 



fore 35,800X4^1 = 26,850 lbs., and the maximum moment of 
flexure at iV = 26,850X60 = 1,61 1,000 in. -lbs. Selecting two 

• 11 IT 11 1611000X7.5 
1 5'mch heavy eye-beams, 7=750X2; c = 7j; .*. / = — 

75^/^2 

= 8055 lbs. = maximum unit stress in the mast. The factor of 
60000 



safety = 



8055 



= 7.4, safe. 



Fig. 273. 




Fig. 274. 

The moment of flexure in the jib is a maximum when the 

maximum load is suspended at its center. The maximum flexure 

PI wP 
moment due to the load and the weight of the jib=«M= — +-^, 

4 8 

in which P=load = 10,000 lbs.; 'ze;= weight of two channels per 
foot of length, =100 lbs. if 12-inch heavy channels are chosen. 
Substituting numerical values, M =49,050 ft.-lbs. = 588,600, in. -lbs. 

Mc 
The resulting maximum fiber stress due to flexure, i = ~r = 



588600X6 
2X248 



= 7120 lbs. 



390 MACHINE DESIGN- 

Compression in jib due to chain tension = 10,000 lbs. ; 

Compression in jib due to load =35,800 lbs.; 

Combined compression due to both = 10,000X35,800 =45,800 lbs. 

Unit compression = combined compression -f- area of the two 

^ 45800 ^ ., 

channels = =1^26 lbs. 

Maximum fiber stress due to combined flexure and com- 

60000 
pression = 7120 + 1526 =8647 lbs. The factor of safety = . 

= 6.96. If a smaller factor of safety were desired, smaller channels 
could be used. The jib may be checked for shear. 

The load might be moved nearly up to the mast, hence the 
joint at F must be designed for a total shear of 10,000 lbs. The 
pin and bearing at G, as well as the supporting framework for 
the bearing must be capable of sustaining a lateral force = jBC 
= 8950 lbs. in any direction. The pivot and step at H must be 
capable of sustaining the lateral force ^^ = 8950 lbs., as well as 
a vertical downward thrust of 10,000 lbs. +the weight of the crane. 

239. Pillar-crane Frame. — Fig. 275 shows an outline of the 
frame of a pillar crane. HN represents the floor level; HK 
represents the pillar, which is extended for support to Z; KM 
represents one or more tension rods; MH represents the brace. 
The load hangs from M in the line ah. The pillar is supported 
horizontally at H^ and vertically and horizontally at L. The 
force polygon shows the horizontal forces at H and Z, = 30,000 lbs., 
and the vertical force at Z= 10,000 lbs. From these the sup- 
ports may be designed. These supports should provide for 
rotary motion of the crane about KL, the axis of the pillar. 
The brace may be treated as in the jib crane, the compressive 
force being 21,400 lbs. The tension rods may be designed for 
the force 15,800 lbs. 

The forces sustained by the pillar are as follows (see Fig. 
276): FEy the horizontal component of i4£, = 15,000 lbs., acts 



MACHINE FRAMES. 



391 



horizontally toward the right at K, and EF, the horizontal 
component of EB, = 15,000 lbs. acts toward the left at H. BC = 
30,000 lbs. acts toward the left at H, and DA =30,000 lbs. acts 
toward the right at L. CD acts upward at L, producing a total 
compression in the portion LH of 10,000 lbs. The force AF = 
5000 lbs. acts to produce tension between H and K. From 
these data the pillar may be designed by methods already given. 

Fig. 275. 




Fig. 276. 



240. Frame of a Steam Riveter. — Let Fig. 277 represent a 
steam riveter. Both the frame and the stake are acted upon by 
three parallel forces when a rivet is being driven. The lines of 
action of these forces AB^ BCy and CA, are ab, be, ca. The 
force AB required to drive the rivet = 35,000 lbs. BC and CA 
may be found, the distances EH and HG being known. The 
moment of flexure on the line ca = 35,000 lbs. X 74'' = 2,590,000 
inch-pounds. Let the line HF represent this moment. The moment 



392 



MACHINE DESIGN. 



in any horizontal cross-section may be found from the diagram 
EFG. Any section of the frame or stake may therefore be checked 
The stake needs to be small as possible in order that small boiler 
shells and large flues may be riveted. In order that it may be 
of equal strength, with the cast iron frame, it is made of material 
of greater unit strength, as cast steel. 




Fig. 277. 

The two bolts which hold the frame and stake together sus- 
tain a force of 107,000 lbs. The force upon each therefore is 
53,500 lbs. If the unit strength of the material is 50,000 lbs., 
and the factor of safety is 6, the area of cross-section of each bolt 

6X53500 
would be = — ; — = 6.42 square inches. The diameter corre- 



50000 
sponding = 2.86 inches. 



A 3i-inch bolt has a diameter at the 



MACHINE FRAMES. 393 

bottom of the thread = 2.88 inches, and and hence 3J-mch bolts 
will serve as far as strength is concerned. But the body of the 
bolt is 60 inches long, and each inch of this length will yield 
a certain amount, and the total yielding might exceed an 
allowable value, even if a safe stress were not exceeded. The 
yielding per inch of length, or the unit strain = umi stress -7- 
coefficient of elasticity, or 

X = -L, but / = J =^ = 6440 lbs. 
and £ = 28,000,000 

•% ^=—^ = .00002^ inch. 

28000000 ^ 

Total yielding = ^ X 60 = .00138 inch. 

This amount of yielding is allowable and therefore two 3J-inch 

bolts will serve. 



APPENDIX. 



The following method of determining the position of the 
slider-crank chain corresponding to the maximum velocity of the 
slider is largely due to Professor L. M. Hoskins. 

Refer to Fig. 278. 




L=2a- 



FlG. 278. 



BM = b = connecting-rod length, to scale. 

OM = a = crank length, to scale. 

OA=y. 

AM = x. 

OB = d. 

Angle AOB =gd^. 

'' MAO = a. 

'' AMO = r. 

'* OMB = ^. 
Z = 2a = length of stroke of slider. 

From our study of the velocity diagram of the slider-crank 
chain (see § 22) we know that the length y will represent the 

395 



39 6 APPENDIX. 

velocity of the slider on the same scale as the length a repre- 
sents the velocity of the center of the crank-pin. The length 
y is determined by erecting at O a perpendicular to the Hne of 
action of the slider and cutting this perpendicular by the con- 
necting-rod b, extended if necessary. 

Our problem, then, is to find the position of the mechanism 
corresponding to the maximum value of y. 

Consider the triangle whose sides are y, Xy and a. Calling 
the angle included between x and y, a, 

a^=y^+x^ — 2xy cosa; ..... (i) 



y 

but COSQ' = -^* ....... (2) 

x + b ^ ^ 



:. a^=y^+x^-2xy^ (3) 



Clearing (3) of fractions and transposing, 

2xy^ = (x^ + y^ — a^) (x + b), 
2xy^=x^+x^b+y^x+y^—a^x—a^b, ... (4) 

Differentiating, 



dy dy dy 

2y^+4xy-^ =3x^+2bx+y^ + 2xy^ +2by^ -a\ 



Transposing, 



dy 
{/^y—2xy—2by)-T- = 2,x'^ -\-2bx-\-y'^ —a^ —2y^. . . (5) 



APPENDIX. 397 

dy 
For maximum value oi y, -i- =o\ hence we may write o for 

the left-hand term of (5). 

.*. ;y2= 3^2 + 26^ — ^2 (6) 

Adding x^ and subtracting a^ from both sides of (6), 

x^+y'^ — a?=/\x^-{-2hx — 2a'^ (7) 



From (3), 



2,2 2 ^^y 



x+y 

Substituting this value in (7), 

qxv 

— fT- = 4X^ + 2bx — 2a^; 

x + b ^ * 

.'. 2xy^ = (4X^ + 2bx — 2a^)(x+b) (8) 

Substituting in (8) the value of y^ given in (6), 

2:^(3x2 + 2bx — a^)= (4^2 -\-2bx — 2^2) (x+b); 
.*. 6x^ + 4bx^ — 2a^x = 4.x^ + 46^2 + 2 6:^2 -|_ 2 ^2^,. _ 2 a^x — 2 Ja^ ; 
2X^ — 2bx^ — 2b^x + 2ba^ =0; 
x^ — bx^ — b^x + &a^^ = o. . . • . . . (9) 

Dividing (9) by a^ and transposing, 

a^ X x^ x^ 

P^6"^62~63 (10) 

Equation (10) gives us the relation existing between a, b, 
and X for the maximum velocity of the slider. 



398 APPENDIX. 

X 

By taking a series of values of 7- and solving (10) for the 

a 
corresponding values of 7-, Curve A has been constructed. Ordi- 

X ^ . a 

nates are t", abscissae are r* , 

For any given problem the values of a and h are known. 

Solve for r. 


X 

From Curve A find the value of 7- corresponding to this value 
of^. 

X 

From the determined value of 7- and the known value of h 



the numerical value of x is found. 

But equation (6) gives for the maximum value of y the rela- 
tion 

y2=2,x'^j^2hx — a^, 

which we can readily solve for y since all of the right-hand terms 
are now known. 

AOB being a right-angled triangle, 

d'^ = {x + hy-y^. 

The values of the right-hand member being known we can 
readily solve this for d. 

Let m represent the distance moved through by the slider 
from the beginning of the stroke, then 

in = h-\-a — d. 



y4PPENDIX. 399 

The portion of the stroke accomplished by the sHder at the 
time of its maximum velocity expressed as a fraction of the 
whole stroke, 2a, 

m 
2a 

Curve B shows the relation between 7- and — . From this 

2a 

a 
curve we can see at a glance for any given value of t what per 

cent of the slider's stroke is accomplished when its position of 

a 
maximum velocity is reached. Abscissae are values of 7- ; ordi- 

m 
nates — . 
2a 

y 

— = ratio of the maximum velocity of the slider to the velocity 

of the center of the crank-pin. Curve C shows the relation 

ay a , 

between the values of t and — . Abscissae are values of 7- : ordi- 



a 



y y 

nates 1. Add unity to the ordinates for actual values of — . 

a ^ a 

To find the values of /? corresponding to the maximum velocity 

of the slider we have the three sides of the triangle OMBy namely, 

h, a, and d. Let 

h + a + d 



s(s — d) 
Then cos J^ = —7 — , from which we can readily get the 

value of ^. 

Curve D is plotted with values of ^, in degrees, as ordinates 

a 

and values of r as abscissae. 





400 



APPENDIX. 



Table XXXI. — ^Assumed Va.lues of — , and Corresponding Computed Values 



OF -r, ^, — , — , AND B, TO Plot Curves A, B, C, and D. 



X 


O.OIO 


0.015 


0.025 


0.035 


0.050 


0.075 


O.IOO 


0.200 


a 


0.1005 


0.1234 


0.1600 


0.1902 


0.2289 


0.2832 


. 3302 


0.4817 


y 

b 


O.IOI 


0.1243 


0. 1621 


0.1936 


0.2348 


. 2944 


0.3479 


0.5367 


m 
2a 


0.4751 


0.4704 


0.4622 


0.4561 


. 4484 


. 4402 


0.4320 


0.4239 


a 


1.005 


1.0073 


1-0131 


I. 0179 


1.0258 


I -0395 


I -0533 


1.1139 


P 


89° 55' 


89° 49' 


89° 44' 


89° 35' 


89° 26' 


88° 50' 


88° 16' 


85° 17' 



0.300 

0.6025 
0.7I2I 

0.4273 

I.I8I9 

81° 23' 



0.400 

0.7043 

0.8854 

o . 4409 

I. 2571 
76° 41' 



0.500 

o . 7906 
1.0607 

0.4597 
I. 3416 



0.600 

0.8626 
1-2393 

0.4901 

1-4367 
65° 19' 



o. 700 

0.9203 
1.4223 
0-5374 

1-5455 



0.800 

0.9633 
I . 6100 
0.6044 

1-6713 
48° 21' 



0.900 

0.9905 

1.8025 

0.7221 

I. 8198 
35° 8' 



APPENDIX, 



401 




INDEX 



PAGB 

Acceleration, diagrams of 60-67 

Addendum 277 

Angle of action 284-286, 287 

Application, machinery of 4 

Arc of action 284-286, 287 

Axle design 170-173 

Axles, shafts, and spindles 170-180 

Backlash 277 

Ball bearings, see Roller- and Ball-bearings. 

Bearing pressure, allowable for journals 182-183 

Allowable for roller- and ball-bearings 218-220 

Allowable for sliding surface 167-168 

Allowable for thrust- bearings . 195 

Bearings 198-204 

See also Journals; Roller- and Ball-bearings. 

Belts 230-255 

Cone pulley 236-239 

Crowning pulleys , 236 

Design, theory of 239-245 

Distance between shafts for 251 

Driving capacity, variation of 248-251 

Dynamo-belt design 246-248 

Intersecting axes 234-235 

Pump-belt design 245-246 

Rope-drives 251-255 

Shifting, principle of 233 

Transmission of motion by 230-233 

Twist 233-234 

Weight of leather 244 

Bevel-gears 304-310 

Skew 311 

Bolts and screws 1 19-147 

Analysis of screw action 122-125 

Calculation of bolts subject to elongation 132-136 

Calculation of screws for transmission of power 140-147 

403 



404 INDEX. 

PAGE 

Calculation of screws not stressed in screwing up 125-126 

Calculation of screws stressed in screwing up 126-136 

Classification and definitions 1 19-132 

Design of bolts for shock 136-139 

Jam-nuts 139 

U. S. standard threads 120-121 

Wrench-pull 131 

Boxes 198-204 

See also Journals. 
Box pillar, see Supports. 

Brackets 84-85 

See also Supports. 

Cams 47-53 

Cast-iron parts 85-86 

Centro 10-12 

Location of 15-16, 1 7, 20-22 

Centrode 12-13 

Centros of three links 16-17 

Clearance 277 

Clutches 225-229 

See also Couplings. 

Connecting-rod, angularity of 28-29 

Constrained motion 4-8 

Cotters 156-157 

Couplings and clutches 221-229 

Claw or toothed 225-226 

Combination friction and claw 229 

Compression 222-223 

Flange 222 

Flexible 224-225 

Friction 226-229 

Hooke's 224 

Oldham's 224 

Sellar's 223 

Sleeve or muff 221 

Weston friction 228-229 

Cranes, problems in design of 379-391 

Crank-pin, design of 191 

Cross-head pin, design of 192 

Cutting speeds 30 

Cycloidal gears 281-284, 288, 290, 291, 301, 305 

Dynamo, belt design for 246-248 

Efficiency of machines 2 

Elements, pairs of motion 13 

Size o{ 17-18 



INDEX, 405 



PAGE 



Energy, definition of i 

In machines 54~7o 

Law of conservation of i 

Sources of 3 

Feathers, see Keys. 

Fly-wheels 256-273 

Construction of 269-273 

Design, general method 257 

Pump 261-264 

Punching-machine 257-261 

Steam-engine 264 

Stresses in arms 267-269 

Stresses in rims 264-267 

Theory of 256-257 

Force, definition of i 

Force-fits 157-161 

Frames 81-84, 359-393 

Closed 370-372 

Cranes 374-391 

Open-side 359-3 7o 

Punching-machine 359-365 

Riveting-machine 391-393 

Slotting-machine 365-367 

Steam-engine, center-crank 370-372 

Steam-engine, side-crank 367-370 

Steam-hammer 373-379 

See also Supports and Machine parts. 

Friction, coefficient for belts 244 

Coefficient for dry surfaces 228 

Coefiicient for ropes 252 

Clutches 226-229 

Tower's experiments 202-205 

See also Journals; Lubrication, and Sliding Surfaces. 

Gearing, see Toothed Wheels 

Graphite, as lubricant 209 

Indicator mechanisms 42-46 

Crosby 43-44 

Tabor 42 

Thompson 43 

Involute gears 286-288, 289, 290, 293, 301, 306 

Instantaneous center 10-12 

Instantaneous motion 10 



4o6 INDEX. 

PAGE 

Jib-crane 379-390 

Journals 181-209 

Allowable bearing pressure 182-183 

Calculation of, for strength 187-193 

Crank-pin of engine 191 

Cross-head pin of engine 192 

General discussion of 181-182 

Heating of 184-186 

Lubrication of 195-196, 201-209 

Main-journal of engine 188-190 

Materials for,and bearings 186-187 

Proportions of 186 

Thrust 193-197 

Tower's experiments on friction of 202-205 

See also Roller- and Ball-bearings. 

Keys 148-157 

Classification of 148 

Cotters 156-157 

Feathers 154-155 

Kernoul and Barbour 152-153 

Parallel 148 

Roller ratchet 153 

Round taper 155 

Saddle, flat, and angle 152 

Splines i54-i55 

Strength of 154 

Taper 148-149 

Woodruff 151-152 

Lathe, bed 84 

Supports 355-358 

Legs, see Supports. 

Lever-crank chain, location of centros 17 

Velocity, diagram of 26 

Linear velocity 22 

Points in different links 26 

Line-shafts = 177-180 

Linkage 14 

Lubrication, of roller- and ball-bearings 220 

Of rotating surfaces 195-196, 201-209 

Of sliding surfaces 166-169 

Machine cycle 2 

Machine frames, form dictated by stress 81-84 

See also Frames. 



INDEX. 407 

PAGE 

Machine function defined 2 

Machine parts, forms of cast members 85-86 

Proportions of 71-86 

Main journal, design of 188-190 

Mechanism, definition of 15 

Location of centros in compound 19-22 

Motion, chains 14-15 

Constrained 4-8 

Definition of i 

Free 4 

Helical 9 

Instantaneous io~i3 

Kinds of, in machines 8-9 

Plane 8 

Relative 9 

Spheric 9 

Pairs of elements 13 

Parallel motions, see Straight-line motions. 

Passive resistance 5-8 

Pillar crane 390"">; 9^ 

Pitch arc 264 

Pitch circle 276 

Pitch, circular 277 

Diametral 277 

Normal 313 

Planing-machine, bed 83-84, 354, 358 

Lubrication of 169 

Table 82-83 

Prime mover 3 

Pulleys, cone 236-239 

Crowning 236 

Idler or guide 235 

Proper size of 251 

Stresses in arms 267-269 

Stresses in rims 2^4-267 

See also Fly-wheels. 

Pump, belt design for 245-246 

Fly-wheel, design for 261-264 

Punching-machine, fly-wheel design for 257-261 

Frame, design for 359- 365 

Quick-return mechanisms 2>^^2)^ 

Lever-crank quick-return 32-35 

Slider-crank quick-return 32-30 

Whitworth quick-return 35~38 



4o8 INDEX. 

PAGE 

Rigid body lo 

Riveted joints 87-118 

Boiler-shell problem 112-118 

Construction for tightness 108-109 

Countersunk rivets 107 

Dimensions of rivet-heads 106 

Efficiencies of various kinds of , 94-102, 103-104 

Failure of 91-92 

General formulae 101-1C2 

Kinds of 90-91 

Length of rivet 106-107 

Margin . . . 103 

Methods of riveting 87-88 

More than two plates iio-iii 

Nickel-steel rivets 107 

Perforation of plate 88-90 

Plates not in same plane 111-112 

Plates with upset edges no 

Slippage of 104-106 

Strength of materials used in 92-94, 109 

Strength, proportions and efficiency of 94-104 

Riveting-machine, action of 87 

Frame design for 39i~393 

Roller- and ball-bearings = „ 210-220 

Allowable loading 218-220 

Forms of races . . , 210-218 

Lubrication and sealing 220 

Rolling, sliding, and spinning 210-217 

Size of 219 

Rope transmission 251-255 

Screws, see Bolts and screws. 

Screw-threads, see Bolts and screws. 

Set-screws . . 120, 157 

Shafting, angular distortion of 176 

Combined thrust and torsion 176-177 

Combined torsion and bending 1 74-375 

Hollow vs. solid 175-176 

Line-shafts 177-180 

Simple torsion 1 73-1 74 

Shaping-machine, force problem 5^5^ 

Quick- return mechanism for 35 

Sheave- wheels, angle of groove , 254 

Diameter of 252 

Shrink-fits 157-160 

Skew bevel-gears 311 



INDEX, 409 

PAGB 

Slider-crank chain, acceleration diagrams 60-67 

Description of 19 

Force problem, shaping-machine 56-58 

Force problems, steam-engine 58-70 

Location of centros 15-16 

Maximum velocity of slider 56, 395-401 

Tangential effort diagrams 68-70 

Velocity diagram 24-26 

Sliding surfaces 162-169 

Allowable bearing pressure 167-168 

Form of guides 164-166 

General discussion 162 

Lubrication of 166-169 

Proportions dictated by wear 163-164 

Slotted cross-head 18-19 

Slotting-machine, frame design for 365-367 

Spindles, see Axles, shafts, and spindles. 

Spiral gears 31 1-341 

Axes at 90° 311-327 

Axes at any angle, B 327-341 

Splines, see Keys. 

Springs 349-353 

Cantilever 35o~352 

Coil or spiral 352-353 

Leaf 351-353 

Spur-gears 275-304, 341-348 

Steam-engine, boxes 199 

Crank-pin 183, 186, 191 

Cross-head pin 183, 186, 192 

Fly-wheel design 264 

Force problems 58-70 

Frame, center-crank 370-372 

Frame, " girder-bed " 367-369 

Frame, " heavy duty " 369-370 

Main journal 183, 186, 188-190 

Steam-hammer, double-acting, frame design 377-379 

Single-acting, frame design 373-377, 378 

Straight-line motions 39-46 

General methods of design , 41-46 

Grasshopper 41-42 

Parallelogram 39-41 

Watt parallel motion 39 

Stresses in machine parts, compression 78-79 

Constant 72 

Flexure 79-80 

Shock 77-78 



4IO INDEX, 

PAGB 

Tension 78 

Torsion 80-81 

Variable 73-77 

Supports 354-358 

Divided 358-361 

General laws for design of 354 

Reduced nuniber of 357-358 

Three-point 35^-357 

See also Brackets, and Frames. 

Thrust-journals 193-197 

Toothed wheels or gears 275-348 

Addendum 277 

Angle of action 284-286 

Annular 290 

Arc of action 284-286 

Backlash 277 

Bevel-gears 304-311 

Circular pitch 277 

Clearance 277 

Cycloidal teeth 281-284, 288, 290, 291, 301, 305 

Depth, total 277 

Depth, working 277 

Diametral pitch 277 

Epicyclic trains 344-348 

Forms of teeth 278-284, 286-288, 294-297 

Interchangeable sets 291-294 

Involute teeth 286-288, 289, 290, 293, 301, 306 

Line of pressure 284 

Non-circular wheels 303-304 

Pinion '. 288 

Pitch arc 289 

Pitch circle 276 

Proportions of 297-298 

Racks 288-289 

Reverted trains 344 

Skew bevel-gears 311 

Spiral gears 31 1-341 

Spur gear-chains, compound 341-348 

Spur-wheels 275-304 

Strength of teeth 299-303, 309, 340 

Theory of 275-276 

Worms and wheels 334-341 

Torque diagrams 68-70 

Transmission, machinery of 3 

See also Belts, Ropes, Shafting, and Toothed Wheels. 



INDEX, 4" 

PAGS 

Vector quantity 23 

Velocity, angular 22 

Linear 22 

Relative 22, 24, 26-27 

Whitworth quick-return mechanism 35-38 

Work, definition of i 

Worm-gearing 334-341 



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Wait's Engineering and Architectural Jurisprudence 8vo, $6 00 

Sheep, 6 50 

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* Greene's Structural Mechanics Svo, 

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HoUey and Ladd's Analysis of Mixed Paints, Color Pigments and Varnishes. 

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Keep's Cast Iron Svo, 

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Martens's Handbook on Testing Materials. (Henning.) 2 vols Svo, 

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Metcalf 's Steel. A Manual for Steel-users 12mo, 

Morrison's Highway Engineering Svo, 

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Richardson's Modern Asphalt Pavements 8vo, 

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Butts's Civil Engineer's Field-book 16mo, mor. 2 50 

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Ives and Hilts's Problems in Surveying, Railroad Surveying and Geodesy 

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* Orrock's Railroad Structures and Estimates Svo, 3 00 

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DRAWING. 

Barr's Kinematics of Machinery 8vo, 

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* " " " Abridged Ed 8vo, 

Coolidge's Manual of Drawing Svo, paper, 

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Hill's Text-book on Shades and Shadows, and Perspective Svo, 

Jamison's Advanced Mechanical Drawing Svo, 

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Jones's Machine Design: 

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Kimball and Barr's Machine Design. (In Press.) 

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Robinson's Principles of Mechanism Svo, 

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Smith (A. W.) and Marx's Machine Design Svo, 

Smith's (R. S.) Manual of Topographical Drawing. (McMillan) Svo, 

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Weisbach's Kinematics and Power of Transmission. (Hermann and 
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Free-hand Lettering .Svo, 

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Benjamin's History of Electricity Svo, 

Voltaic Cell Svo, 

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Betts's Lead Refining and Electrolysis 8vo, $4 00 

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Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 4 00 

Flather's Dynamometers, and the Measurement of Power 12mo, 3 00 

Getman's Introduction to Physical Science 12mo, 

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Kinzbrunner's Testing of Continuous-current Machines Svo, 2 00 

Landauer's Spectrum Analysis. (Tingle.) Svo, 3 00 

Le Chatelier's High-temperature Measurements. (Boudouard — Burgess. )12mo, 3 00 

Lob's Electrochemistry of Organic Compounds. (Lorenz) Svo, 3 00 

* Lyndon's Development and Electrical Distribution of Water Power. .Svo, 3 00 

* Lyons's Treatise on Electromagnetic Phenomena. Vols, I .and II. Svo, each, 6 00 

* Michie's Elements of Wave Motion Relating to Sound and Light Svo, 4 OU 

Morgan's Outline of the Theory of Solution and its Results 12mo, 1 00 

* Physical Chemistry for Electrical Engineers 12mo, 1 50 

* Norris's Introduction to the Study of Electrical Engineering Svo, 2 50 

Norris and Dennison's Course of Problems on the Electrical Characteristics of 

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* Parshall and Hobart's Electric Machine Design 4to, half mor, 12 50 

Reagan's Locomotives: Simple, Compound, and Electric. New Edition. 

Large 12mo, 3 50 

* Rosenberg's Electrical Engineering. (Haldane Gee — Kinzbrunner.) . .8vo, 2 00 

Ryan, Norris, and Hoxie's Electrical Machinery. Vol. I Svo, 2 50 

Schapper's Laboratory Guide for Students in Physical Chemistry 12mo, 1 00 

* Tillman's Elementary Lessons in Heat Svo, 1 50 

Tory and Pitcher's Manual of Laboratory Physics Large 12mo, 2 00 

Ulke's Modern Electrolytic Copper Refining Svo, 3 00 



LAW. 

* Brennan's Hand-book of Useful Legal Information for Business Men. 

16mo, mor. 5 00 

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Manual for Courts-martial 16mo, mor. 1 50 

Wait's Engineering and Architectural Jurisprudence Svo, 6 00 

Sheep, 6 50 

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Sheep, 5 50 



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Baker's Elliptic Fvinctions Svo, 1 50 

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* Buchanan's Plane and Spherical Trigonometry Svo, 1 00 

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Byerley's Harmonic Functions gvo, 

Chandler's Elements of the Infinitesimal Calculus 12mo, 

* Coffin's Vector Analysis 12mo, 

Compton's Manual of Logarithmic Computations 12mo, 

* Dickson's College Algebra Large 12mo, 

* Introduction to the Theory of Algebraic Equations Large 12mo, 

Emch's Introduction to Projective Geometry and its Application 8vo, 

Fiske's Functions of a Complex Variable 8vo, 

Halsted's Elementary Synthetic Geometry 8vo, 

Elements of Geometry Svo, 

* Rational Geometry 12mo, 

Hyde's Grassmann's Space Analysis Svo, 

♦Johnson's (J. B.) Three-place Logarithmic Tables : Vest-pocket size, paper, 

100 copies, 
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10 copies, 
Johnson's (W. W.) Abridged Editions of Differential and Integral Calculus. 

Large 12mo, 1 vol. 

Curve Tracing in Cartesian Co-ordinates 12mo, 

Differential Equations Svo, 

Elementary Treatise on Differential Calculus Large 12mo, 

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Treatise on Differential Calculus Large 12mo, 

Treatise on the Integral Calculus Large 12mo, 

Treatise on Ordinary and Partial Differential Equations. . .Large 12mo, 

Karapetoff's Engineering Applications of Higher Mathematics. 

(In Preparation.) 
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* Ludlow and Bass's Elements of Trigonometry and Logarithmic and Other 

Tables Svo, 3 00 

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Macfarlane's Vector Analysis and Quaternions Svo, 1 00 

McMahon's Hyperbolic Functions Svo, 1 00 

Manning's Irrational Numbers and their Representation by Sequences and 

Series 12mo, 1 25 

Mathematical Monographs. Edited by Mansfield Merriman and Robert 

S. Woodward Octavo, each 1 00 

No. 1. History of Modem Mathematics, by David Eugene Smith. 
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No. 3. Determinants, by Laenas Gifford Weld. No. 4. Hyper- 
bolic Functions, by James McMahon. No. 5. Harmonic Func- 
tions, by William E. Byerly. No. 6. Grassmann's Space Analysis, 
by Edward W. Hyde. No. 7. Probability and Theory of Errors, 
by Robert S. Woodward. No. S. Vector Analysis and Quaternions, 
by Alexander Macfarlane. No. 9. Differential Equations, by 
William Woolsey Johnson. No. 10. The Solution of Equations, 
by Mansfield Merriman. No. 11. Functions of a Complex Variable, 
by Thomas S. Fiske. 

Maurer's Technical Mechanics Svo, 4 00 

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Solution of Equations Svo, 1 00 

Rice and Johnson's Differential and Integral Calculus. 2 vols, in one. 

Large 12mo, 1 50 

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Smith's History of Modern Mathematics Svo, 1 00 

* Veblen and Lennes's Introduction to the Real Infinitesimal Analysis of One 

Variable Svo, 2 00 

* Waterbury's Vest Pocket Hand-book of Mathematics for Engineers. 

2i X 5| inches, mor. 1 00 

Weld's Determinants Svo, 1 00 

Wood's Elements of Co-ordinate Geometry Svo, 2 00 

Woodward's Probability and Theory of Errors Svo, 1 00 



12 



MECHANICAL ENGINEERING. 

MATERIALS OF ENGINEERING. STEAM-ENGINES AND BOILERS. 

Bacon's Forge Practice 12mo, 

Baldwin's Steam Heating for Buildings 12mo, 

Barr's Kinenvatics of Machinery 8vo, 

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* " " " Abridged Ed 8vo. 

* Burr's Ancient and Modem Engineering and the Isthmian Canal 8vo, 

Carpenter's Experimental Engineering 8vo, 

Heating and Ventilating Buildings 8vo, 

Clerk's Gas and Oil Engine. (New edition in press.) 

Compton's First Lessons in Metal Working 12mo, 

Compton and De Groodt's Speed Lathe 12mo, 

Coolidge's Manual of Drawing 8vo, paper, 

Coolidge and Freeman's Elements of Geenral Drafting for Mechanical En- 
gineers Oblong 4to, 

Cromwell's Treatise on Belts and Pulleys 12mo, 

Treatise on Toothed Gearing 12mo, 

Dingey's Machinery Pattern Making 12mo, 

Durley's Kinematics of Machines 8vo, 

Flanders's Gear-cutting Machinery Large 12mo, 

Flather's Dynamometers and the Measurement of Power 12mo, 

Rope Driving 12mo, 

Gill's Gas and Fuel Analysis for Engineers 12mo, 

Goss's Locomotive Sparks 8vo, 

Greene's Pumping Machinery. (In Preparation.) 

Hering's Ready Reference Tables (Conversion Factors) 16mo, mor. 

* Hobart and Ellis's High Speed Dynamo Electric Machinery 8vo, 

Hutton's Gas Engine 8vo, 

Jamison's Advanced Mechanical Drawing 8vo, 

Elements of Mechanical Drawing 8vo, 

Jones's Gas Engine 8vo, 

Machine Design: 

Part I. Kinematics of Machinery 8vo, 

Part II. Form, Strength, and Proportions of Parts 8vo, 

Kent's Mechanical Engineer's Pocket-Book 16mo, mor. 

Kerr's Power and Power Transmission 8vo, 

Kimball and Barr's Machine Design. (In Press.) 

Levin's Gas Engine. (In Press.) 8vo, 

Leonard's Machine Shop Tools and Methods 8vo, 

* Lorenz's Modem Refrigerating Machinery. (Pope, Haven, and Dean)..8vo, 
MacCord's Kinematics; or. Practical Mechanism 8vo, 

Mechanical Drawing 4to, 

Velocity Diagrams 8vo, 

MacFarland's Standard Reduction Factors for Gases 8vo, 

Mahan's Industrial Drawing. (Thompson.) 8vo, 

Mehrtens's Gas Engine Theory and Design Large 12mo, 

Oberg's Handbook of Small Tools Large 12mo, 

* Parshall and Hobart's Electric Machine Design. Small 4to, half leather, 

Peele's Compressed Air Plant for Mines 8vo, 

Poole's Calorific Power of Fuels 8vo, 

* Porter's Engineering Reminiscences, 1855 to 1882 8vo, 

Reid's Course in Mechanical Drawing 8vo, 

Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 

Richards's Compressed Air 12mo, 

Robinson's Principles of Mechanism Svo, 

Schwamb and Merrill's Elements of Mechanism Svo, 

Smith (A. W.) and Marx's Machine Design Svo, 

Smith's (O.) Press- working of Metals Svo, 

Sorel's Carbureting and Combustion in Alcohol Engines. (Woodward and 

Preston.) Large 12mo, 

Stone's Practical Testing of Gas and Gas Meters Svo, 

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Thurston's Animal as a Machine and Prime Motor, and the Laws of Energetics. 

12mo, $1 00 

Treatise on Friction and Lost Work in Machinery and Mill Work. . .8vo, 3 00 

* Tillson's Complete Automobile Instructor 16mo, 1 50 

* Titsworth's Elements of Mechanical Drawing Oblong 8vo, 1 25 

Warren's Elements of Machine Construction and Drawing 8vo, 7 50 

* Waterbury's Vest Pocket Hand-book of Mathematics for Engineers. 

2|X5f inches, mor. 1 00 
Weisbach's Kinematics and the Power of Transmission. (Herrmann — 

Klein.) Svo, 5 00 

Machinery of Transmission and Governors. (Hermann — Klein.). .Svo, 5 00 

Wood's Turbines 8vo, 2 50 



MATERIALS OF ENGINEERING. 

* Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 

Burr's Elasticity and Resistance of the Materials of Engineering Svo, 7 50 

Church's Mechanics of Engineering Svo, 6 00 

* Greene's Structural Mechanics Svo, 2 50 

* HoUey's Lead and Zinc Pigments Large 12mo 3 00 

HoUey and Ladd's Analysis of Mixed Paints, Color Pigments, and Varnishes. 

Large 12mo, 2 50 
Johnson's (CM.) Rapid Methods for the Chemical Analysis of Special 

Steels, Steel-Making Alloys and Graphite Large 12mo, 3 00 

Johnson's (J. B.) Materials of Construction Svo, 

Keep's Cast Iron Svo, 

Lanza's Applied Mechanics Svo, 

Maire's Modern Pigments and their Vehicles 12mo, 

Martens's Handbook on Testing Materials. (Henning.) SvO; 

Maurer's Techincal Mechanics Svo, 

Merriman's Mechanics of Materials Svo, 

* Strength of Materials 12mo, 

Metcalf's Steel. A Manual for Steel-users 12mo, 

Sabin's Industrial and Artistic Technology of Paint and Varnish Svo, 

Smith's ((A. W.) Materials of Machines 12mo, 

Smith's (H. E.) Strength of Material 12mo. 

Thurston's Materials of Engineering 3 vols., Svo, 

Part I. Non-metallic Materials of Engineering, Svo, 

Part II. Iron and Steel Svo, 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents Svo, 

Wood's (De V.) Elements of Analytical Mechanics Svo, 

Treatise on the Resistance of Materials and an Appendix on the 

Preservation of Timber Svo, 2 00 

Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and 

Steel Svo, 4 00 



STEAM-ENGINES AND BOILERS. 

Berry's Temperature-entropy Diagram 12mo, 2 00 

Carnot's Reflections on the Motive Power of Heat. (Thurston.) 12mo, 1 50 

Chase's Art of Pattern Making 12mo, 2 50 

Creigh ton's Steam-engine and other Heat Motors Svo, 5 00 

Dawson's "Engineering" and Electric Traction Pocket-book. .. . 16mo, mor. 5 00 

Ford's Boiler Making for Boiler Makers ISmo, 1 00 

* Gebhardt's Steam Power Plant Engineering Svo, 6 00 

Goss's Locomotive Performance Svo, 5 00 

Hemenway's Indicator Practice and Steam-engine Economy 12mo. 2 00 

Hutton's Heat and Heat-engines Svo, 5 00 

Mechanical Engineering of Power Plants Svo, 5 00 

Kent's Steam boiler Economy Svo, 4 00 

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Kneass's Practice and Theory of the Injector 8vo, $1 50 

MacCord's Slide-valves 8vo, 2 00 

Meyer's Modern Locomotive Construction 4to, 10 00 

Moyer's Steam Turbine 8vo. 4 00 

Peabody's Manual of the Steam-engine Indicator 12mo. 1 50 

Tables of the Properties of Steam and Other Vapors and Temperature- 
Entropy Table. . . . ^. Svo. 1 00 

Thermodynamics of the'Steam-engine and Other Heat-engines. . . .Svo, 5 00 

Valve-gears for Steam-engines Svo, 2 50 

Peabody and Miller's Steam-boilers Svo, 4 00 

Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors. 

(Osterberg.) 12mo. 1 25 

Reagan's Locomotives: Simple, Compound, and Electric. New Edition. 

Large 12mo, 3 50 

Sinclair's Locomotive Engine Running and Management 12mo. 2 00 

Smart's Handbook of Engineering Laboratory Practice 12mo, 2 50 

Snow's Steam-boiler Practice Svo, 3 00 

Spangler's Notes on Thermodynamics 12mo, 1 00 

Valve-gears Svo, 2 50 

Spangler, Greene, and Marshall's Elements of Steam-engineering Svo 3 00 

Thomas's Steam-turbines Svo. 4 00 

Thurston's Handbook of Engine and Boiler Trials, and the Use of the Indi- 
cator and the Prony Brake Svo, 5 00 

Handy Tables Svo, 1 50 

Manual of Steam-boilers, their Designs, Construction, and Operation Svo, 5 00 

Manual of the Steam-engine 2vols., Svo. 10 00 

Part I. History, Structure, and Theory. Svo, 6 00 

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Steam-boiler Explosions in Theory and in Practice 12mo, 1 50 

Wehrenfenning's Analysis and Softening of Boiler Feed-water. (Patterson). 

Svo. 4 00 

Weisbach's Heat, Steam, and Steam-engines. (Du Bois.) Svo. 5 00 

Whitham's Steam-engine Design Svo, 5 00 

Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .Svo, 4 00 



MECHANICS PURE AND APPLIED. 

Church's Mechanics of Engineering Svo. 6 00 

Notes and Examples in Mechanics Svo, 2 00 

Dana's Text-book of Elementary Mechanics for Colleges and Schools .12mo, 1 50 
Du Bois's Elementary Principles of Mechanics: 

Vol. I. Kinematics Svo. 3 50 

Vol. II. Statics. Svo. 4 00 

Mechanics of Engineering. Vol. I Small 4to, 7 50 

Vol. II Small 4to, 10 00 

* Greene's Structural Mechanics Svo, 2 50 

James's Kinematics of a Point and the Rational Mechanics of a Particle. 

Large 12mo. 2 00 

* Johnson's (W. W.) Theoretical Mechanics 12mo, 3 00 

Lanza's Applied Mechanics Svo. 7 50 

* Martin's Text Book on Mechanics. Vol. I, Statics I2mo, 1 25 

* Vol. II, Kinematics and Kinetics. 12mo, 1 50 

Maurer's Technical Mechanics Svo, 4 00 

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Mechanics of Materials Svo, 5 00 

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Robinson's Principles of Mechanism Svo, 3 00 

Sanborn's Mechanics Problems Large 12mo, 1 50 

Schwamb and Merrill's Elements of Mechanism Svo, 3 00 

Wood's Elements of Analytical Mechanics Svo, 3 00 

Principles of Elementary Mechanics 12mo. 1 25 



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MEDICAL. 

* Abderhalden's Physiological Chemistry in Thirty 'Lectures. (Hall and 

Defren.) 8vo, 

von Behring's Suppression of Tuberculosis. (Bolduan.) 12mo, 

Bolduan's Immune Sera 12mo, 

Bordet's Studies in Immunity. (Gay). (In Press.) 8vo, 

Davenport's Statistical Methods with Special Reference to Biological Varia- 
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Ehrlich's Collected Studies on Immunity. (Bolduan.) 8vo, 

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de Fursac's Manual of Psychiatry. (Rosanofi and Collins.).. . .Large 12mo, 

Hammarsten's Text-book on Physiological Chemistry. (Mandel.) 8vo, 

Jackson's Directions for Laboratory Work in Physiological Chemistry . .8vo, 

Lassar-Cohn's Practical Urinary Analysis. (Lorenz.) 12mo, 

Mandel's Hand-book for the Bio-Chemical Laboratory 12mo, 

* Pauli's Physical Chemistry in the Service of Medicine. (Fischer.) ..12mo, 

* Pozzi-Escot's Toxins and Venoms and their Antibodies. (Cohn.). . 12mo, 

Rostoski's Serum Diagnosis. (Bolduan.) 12mo, 

Ruddiman's Incompatibilities in Prescriptions 8vo, 

Whys in Pharmacy 12mo, 

Salkowski's Physiological and Pathological Chemistry. (Omdorff.) .. ..8vo, 

* Satterlee's Outlines of Human Embryology 12mo, 

Smith's Lecture Notes on Chemistry for Dental Students 8vo, 

* Whipple's Tyhpoid Fever Large 12mo, 

WoodhuU's Notes on MiUtary Hygiene 16mo, 

* Personal Hygiene 12mo, 

Worcester and Atkinson's Small Hospitals Establishment and Maintenance, 
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Betts's Lead Refining by Electrolysis 8vo, 4 00 

BoUand's Encyclopedia of Founding and Dictionary of Foundry Terms used 

in the Practice of Moulding 12mo, 

Iron Founder 12mo, 

Supplement 12mo, 

Douglas's Untechnical Addresses on Technical Subjects 12mo, 

Goesel's Minerals and Metals: A Reference Book 16mo, mor. 

* Iles's Lead-smelting 12mo, 

Johnson's Rapid Methods for the Chemical Analysis of Special Steels, 

Steel-making Alloys and Graphite Large 12mo, 

Keep's Cast Iron 8vo, 

Le Chatelier's High- temperature Measurements. (Boudouard — Burgess.) 

12mo, 

Metcalf 's Steel. A Manual for Steel-users 12mo. 

Minet's Production of Aluminum and its Industrial Use. (Waldo.). . 12mo, 

Ruer's Elements of Metallography. (Mathewson) 8vo. 

Smith's Materials of Machines 12mo, 

Tate and Stone's Foundry Practice 12mo, 

Thurston's Materials of Engineering. In Three Parts 8vo, 8 00 

Part I. Non-metallic Materials of Engineering, see Civil Engineering, 
page 9. 

Part II. Iron and Steel 8vo, 3 50 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents 8vo, 2 50 

Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 

West's American Foundry Practice 12mo, 2 50 

Moulders' Text Book 12mo, 2 50 

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MINERALOGY. 

Baskerville's Chemical Elements, (In Preparation.). 

Boyd's Map of Southwest Virginia Pocket-book form. $2 00 

* Browning's Introduction to the Rarer Elements 8vo, 

Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 

Butler's Pocket Hand-book of Minerals 16mo, mor. 

Chester's Catalogue of Minerals 8vo, paper, 

Cloth, 

* Crane's Gold and Silver 8vo, 

Dana's First Appendix to Dana's New " System of Mineralogy" . . Large 8vo, 
Dana's Second Appendix to Dana's New "System of Mineralogy." 

Large 8vo, 

Manual of Mineralogy and Petrography 12mo, 

Minerals and How to Study Them 12mo, 

System of Mineralogy Large 8vo, half leather. 

Text-book of Mineralogy 8vo, 

Douglas's Untechnical Addresses on Technical Subjects 12mo, 

Eakle's Mineral Tables 8vo, 

Eckel's Stone and Clay Products Used in Engineering. (In Preparation). 

Goesel's Minerals and Metals: A Reference Book 16mo, mor. 

Groth's Introduction to Chemical Crystallography (Marshall) 12mo, 

* Hayes's Handbook for Field Geologists 16mo, mor. 

Iddings's Igneous Rocks 8vo, 

Rock Minerals 8vo, 

Johannsen's Determination of Rock-forming Minerals in Thin Sections. 8vo, 

With Thumb Index 

* Martin's Laboratory Guide to Qualitative Analysis with the Blow- 

pipe 12mo, 

Menill's Non-metallic Minerals: Their Occurrence and Uses 8vo, 

Stones for Building and Decoration 8vo, 

* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo, paper, 50 
Tables of Minerals, Including the Use of Minerals and Statistics of 

Domestic Production 8vo, 1 00 

* Pirsson's Rocks and Rock Minerals ^ 12mo, 2 50 

* Richards's Synopsis of Mineral Characters 12mo, mor. 1 25 

* Ries's Clays: Their Occurrence, Properties and Uses 8vo, 5 00 

* Ries and Leighton's History of the Clay-working Industry of the United 

States 8vo, 2 50 

* Tillman's Text-book of Important Minerals and Rocks 8vo, 2 00 

Washington's Manual of the Chemical Analysis of Recks. . ,,,,,,,,.. .Svo, 2 00 



MINING. 

* Beard's Mine Gases and Explosions Large 12mo, 3 00 

Boyd's Map of Southwest Virginia Pocket-book form, 2 00 

* Crane's Gold and Silver 8vo, 5 Of) 

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* 8vo, mor. 5 00 

Douglas's Untechnical Addresses on Technical Subjects 12mo, 1 00 

Eissler's Modern High Explosives 8vo, 4 00 

Goesel's Minerals and Metals: A Reference Book 16mo, mor. 3 00 

Ihlseng's Manual of Mining 8vo, 5 00 

* Iles's Lead Smelting 12mo, 2 50 

Peele's Compressed Air Plant for Mines 8vo, 3 00 

Riemer's Shaft Sinking Under Difficult Conditions. (Coming and Peel»).8vo, 3 00 

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Wilson's Hydraulic and Placer Mining. 2d edition, rewritten 12moi 2 50 

Treatise on Practical and Theoretical Mine Ventilation 12mo, 1 25 

17 



SANITARY SCIENCE. 

Association of State and National Food and Dairy Departments, Hartford 

Meeting, 1906 8vo, 

Jamestown Meeting, 1907 8vo, 

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Sanitation of a Country House 12mo, 

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Folwell's Sewerage. (Designing, Construction, and Maintenance.) 8vo, 

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Fowler's Sewage Works Analyses 12mo, 

Fuertes's Water-filtration Works 12mo, 

Water and Public Health 12mo, 

Gerhard's Guide to Sanitary Inspections 12mo, 

* Modem Baths and Bath Houses 8vo, 

Sanitation of Public Buildings 12mo, 

Hazen's Clean Water and How to Get It Large 12mo, 

Filtration of Public Water-supplies 8vo, 

Kinnicut, Winslow and Pratt's Purification of Sewage. (In Preparation.) 
Leach's Inspection and Analysis of Food with Special Reference to State 

Control 8vo, 

Mason's Examination of Water. (Chemical and Bacteriological) 12mo, 

Water-supply. (Considered principally from a Sanitary Standpoint). 

8vo, 

* Merriman's Elements of Sanitary Enigneering 8vo, 

Ogden's Sewer Design 12mo, 

Parsons's Disposal of Municipal Refuse 8vo, 

Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- 
ence to Sanitary Water Analysis 12mo, 

* Price's Handbook on Sanitation 12mo, 

Richards's Cost of Cleanness 12mo, 

Cost of Food. A Study in Dietaries 12mo, 

Cost of Living as Modified by Sanitary Science 12mo, 

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* Richards and Williams's Dietary Computer Svo, 1 50 

Richards and Woodman's Air, Water, and Food from a Sanitary Stand- 
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* Richey's Plumbers', Steam-fitters', and Tinners' Edition (Building 

Mechanics' Ready Reference Series). . 16mo, mor. 

Rideal's Disinfection and the Preservation of Food Svo, 

Sewage and Bacterial Purification of Sewage Svo, 

Soper's Air and Ventilation of Subways 12mo, 

Turneaure and Russell's Public Water-supplies Svo, 

Venable's Garbage Crematories in America Svo, 

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Ward and Whipple's Freshwater Biology. (In Press.) 

Whipple's Microscopy of Drinking-water Svo, 

* Typhoid Fever Large 12mo, 

Value of Pure Water Large 12mo, 

Winslow's Systematic Relationship of the Coccaceae Large 12mo, 



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Fitzgerald's Boston Machinist ISmo, 1 00 

Gannett's Statistical Abstract of the World 24mo, 75 

Haines's American Railway Management 12mo, 2 50 

* Hanusek's The Microscopy of Technical Products. (Winton) Svo, 5 00 

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Jacobs's Betterment Briefs. A Collection of Published Papers on Or- 
ganized Industrial Efficiency 8vo, 

Metcalfe's Cost of Manufactures, and the Administration of Workshops.. 8vo, 

Putnam's Nautical Charts 8vo, 

Ricketts's History of Rensselaer Polytechnic Institute 1824-1894. 

Large 12mo, 

Rotherham's Emphasised New Testament Large 8vo, 

Rust's Ex-Meridian Altitude, Azimuth and Star-finding Tables 8vo, 

Standage's Decoration of Wood, Glass, Metal, etc 12mo, 

Thome's Structural and Physiological Botany. (Bennett) 16mo, 

Westermaier's Compendium of General Botany. (Schneider) 8vo, 

Winslow's Elements of Applied Microscopy 12mo, 1 50 



HEBREW AND CHALDEE TEXT-BOOOKS. 

Gesenius's Hebrew and Chaldee Lexicon to the Old Testament Scriptures. 

(Tregelles.) Small 4to, half mor, 5 00 

Green's Elementary Hebrew Grammar 12mo, 1 25 



$3 


50 


5 


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2 


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3 


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5 


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2 


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2 


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2 


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19 



SEP 801909 



SEP 80 



1909 









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LIBRARY OF CONGRESS 



021 225 277 3 



